Advanced Experiments in Physics. Physics 306

Advanced Experiments in Physics.Physics 306 University of Wisconsin-Parkside Jeffrey R. Schmidt October 1997, 1999, 2001, 2003, Revision: December 2005 NaI Photocathode Dynodes 1 Contents 1 Introduction 4 2 Nuclear Counting Statistics 2.1 Apparatus . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Background . . . . . . . . . . . . . . . . . . . . . . . 2.3 Geiger tube calibration . . . . . . . . . . . . . . . . . 2.4 Experiment I; Normal distributions . . . . . . . . . . 2.5 Experiment II. Hypothesis testing and goodness of fit 3 The 3.1 3.2 3.3 3.4 3.5 Half-life of Ba137 Background . . . . . Apparatus . . . . . . Procedure . . . . . . Data and analysis . . Background material; . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . the Nuclear Shell . . . . . . . . . . . . . . . . Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 5 5 6 8 14 . . . . . 20 20 22 23 23 27 4 Determination of the Speed of Light 36 5 Determination of CCvp for a gas 5.1 Background . . . . . . . . . . . . . . . . . . . . 5.2 Apparatus . . . . . . . . . . . . . . . . . . . . . 5.3 Procedure . . . . . . . . . . . . . . . . . . . . . 5.4 Data and analysis . . . . . . . . . . . . . . . . . 5.5 Molecular spectra and the equipartition theorem 5.6 Safe Handling of High Pressure Gas Cylinders . 5.7 Lissojous program . . . . . . . . . . . . . . . . . . . . . . . . 40 40 42 42 44 45 49 49 . . . . . 53 53 54 57 61 64 Photoelectric Effect Apparatus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Background . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Procedure and data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69 69 70 71 6 Electron Diffraction and Bragg Scattering 6.1 Apparatus . . . . . . . . . . . . . . . . . . . . 6.2 Background . . . . . . . . . . . . . . . . . . . 6.3 Data and analysis . . . . . . . . . . . . . . . . 6.4 Crystal structure determination by scattering 6.5 Simulation of scattering . . . . . . . . . . . . 7 The 7.1 7.2 7.3 8 Measuring Planck’s constant (solid state 8.1 Theory . . . . . . . . . . . . . . . . . . . 8.2 Procedure . . . . . . . . . . . . . . . . . 8.2.1 Finding roots by bisection . . . . 8.2.2 Newton’s method . . . . . . . . . 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . version) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73 73 78 79 80 9 Helium Ionization Potential; The Franck-Hertz 9.1 The old version of the experiment; apparatus . . 9.2 Data and analysis . . . . . . . . . . . . . . . . . 9.3 Second version . . . . . . . . . . . . . . . . . . 10 The 10.1 10.2 10.3 10.4 10.5 Rydberg Constant Background . . . . . Atomic Spectra . . . Apparatus . . . . . . Data and analysis . . Diffraction gratings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Experiment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83 84 86 87 89 89 94 98 100 102 11 Measurement of unknown spectral 11.1 The Mercury data . . . . . . . . . 11.2 First order calibration curve . . . 11.3 Measuring Sodium lines . . . . . lines 106 . . . . . . . . . . . . . . . . . . . . . . . . 107 . . . . . . . . . . . . . . . . . . . . . . . . 108 . . . . . . . . . . . . . . . . . . . . . . . . 108 12 Stefan-Boltzmann Radiation Law 12.1 Background . . . . . . . . . . . . 12.2 Luminosity or radiance . . . . . . 12.3 Apparatus . . . . . . . . . . . . . 12.4 Procedure . . . . . . . . . . . . . 12.5 Data and analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109 109 115 116 122 122 13 Measurement of the Compton Edge 13.1 Background . . . . . . . . . . . . . . . . . . 13.2 Apparatus . . . . . . . . . . . . . . . . . . . 13.3 Scintillation counters and photomultipliers . 13.4 Stage 1; calibration . . . . . . . . . . . . . . 13.5 Stage II; photopeak and edge measurements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126 126 129 130 134 135 Electrons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135 136 136 137 14 Energy versus Momentum 14.1 Apparatus . . . . . . . . 14.2 Procedure . . . . . . . . 14.3 Data and analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . for Relativistic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 Michelson-Interferometer and the Index of Refraction of Air 138 16 Angular distribution of emitted radiation 142 16.1 Polarization and angular distributions . . . . . . . . . . . . . . . . . . . . . . 145 16.2 Angular correlations of γ emissions . . . . . . . . . . . . . . . . . . . . . . . 148 16.3 A simulated experiment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149 3 17 A simulated experiment; Geiger-Marsden experiment 17.1 The numerical experiment . . . . . . . . . . . . . . . . . 17.2 Raw data . . . . . . . . . . . . . . . . . . . . . . . . . . 17.3 Problem 1. Analysis of simulated lab data . . . . . . . . 17.4 The Thompson model . . . . . . . . . . . . . . . . . . . 17.5 Monte Carlo scattering simulation . . . . . . . . . . . . . 17.6 Energy dependence in the Geiger-Marsden experiment . 17.7 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . 17.8 Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 Appendix 18.1 Appendix 18.2 Appendix 18.3 Appendix 18.4 Appendix 18.5 Appendix 18.6 Appendix 1 . . . . . . . . . . . . . . . . I. Reformatting Lab Data with Sed and Awk . . . II; Nonlinear Least Squares error matrix method . III. The Plotting of Lab Data with GNU plotutils IV. Preparing a lab report with LaTeX . . . . . . V. The Oscilloscope . . . . . . . . . . . . . . . . . VI. Support software . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154 163 164 165 167 168 170 174 175 . . . . . . 178 178 184 187 191 195 196 Introduction This lab book is a synopsis of the setup, required apparatus, and data collection and analysis for each of our Modern Physics Laboratory experiments. The data for each experiment are real, taken in 1996, 1997, 1999, and 2001. The book can serve as an outline and basic reference for the setup of the experiments and their proper analysis. I do not generally teach this course at the present time, so this has been prepared as a courtesy to the students and any person who must teach the course for the first time, but mostly in preparation for the day when I will be teaching it more regularly. In addition to twelve basic experiments, and a variation on one or two of these, there are a number of simulated experiments; things that would be interesting to do if we had the equipment. These include the Geiger-Mardsen experiment, Laue scattering, and γ − γ angular correlations. There are two numerical labs in the Geiger-Marsden section, one verifying the energy dependence of the scattering cross section, another verifying the Rutherford formula for the number of α particles scattered from a gold foil. Hopefully I will add more simulations as time allows. c 1997, 2003, 2005 Jeffrey R. Schmidt The experiments 4 333 M eV β − .6 year .514 M eV γ.184 M eV γ. 0. 3. We will determine whether or not nuclear decays are random events by counting radioactive decays with a Geiger tube and a scalar-counter.172 M eV γ. Radioactive source.320 M eV β. Scaler-timer. 1. 2. Some of these are given in the table below Isotope 137 .542 M eV γ. P r 144 Emission β − . 0. 0. Geiger tube. 55 Cs 27 Co 58 Ce 60 .1µc Cs137 . 2.134 M eV γ.511 M eV (From β − β + → γγ) 5 Half-life 27 year 5. SARGENT−WELCH | SCALAR \ TIMER Count Geiger Tube Voltage Mode Ext 50/60 Hz Freq Volts Preset Freq Time 2.081 M eV β + . Amerschorn-Searle 184471. 0. 1.312 M eV γ.2 Background In at least four experiments in 306 you will study radiations emitted from unstable isotopes. 0. 0.662 M eV β − . 144 11 N a 22 137 56 Ba 58 N i 60 . 0. 0. 1.2 Nuclear Counting Statistics Nuclear decays are random processes that provide an excellent context in which to study basic statistics and distribution functions.2 year 285 days 2.1 Start Count Time Preset Time 10M 100k 1s 10s 10k 1k 100 10 1min 10min 1hr 10hr Reset Power Apparatus 1. 0. 0. Welch Scientific #1216.277 M eV γγ. Sargent-Welch model S-72095-10. but not before it can be registered on the supporting electronics connected to the tube. The tube is filled with air at a very low pressure. which will in turn be accelerated by the potential difference in the tube. the number of counts blows up as the tube registers spurious counts from dielectric breakdowns. and a coaxial metal cylinder enclosed in the glass. with wire positive and cylinder negative. the tube must be calibrated by counting a constant source of ionizing radiation and generating the characteristic curve of counts versus voltage for the particular tube. This is the data that will be statistically analyzed. The counter is a cylindrical glass tubewith a thin tungsten wire running down the center. causing the net potential difference between wire and cylinder to drop below the critical value needed to sustain the cascade of ions. The back-voltage across R combines with the voltage on the tube and diminishes it. Proper operation of the tube depends critically on the voltage applied to it. As the voltage continues to climb. a few torr. Too high a voltage and the tube discharges spuriously from dielectric breakdown. The center of this plateau is the desired operating voltage. and is counted at a variety of applied voltages. A voltage typically around 1000 V is applied between the wire and cylinder. This causes more collisions. liberating more ions and creating a current flow from wire to cylinder and through the external resistor R. 2. This shuts off the current. as the voltage is increased above this point.3 Geiger tube calibration A constant source is placed about 20 cm from the tube. registering false counts even with no ionizing radiation present. A selection of our Geiger counters is shown below. Tungsten wire cylinder R 1000 V Ionizing radiation entering the tube will free ions from air molecules. For voltages below the threshold value there will be no counts detected by the tube. This voltage is slightly below dielectric breakdown voltage for the gas in the tube.The Geiger tube is used to detect the decay particles emitted by the Cesium sample. the number of registered counts will increase and reach a plateau. 6 . To find the optimal operating voltage. The scalar-timer counts the number of decays from the source in ten second intervals. Voltage (volts) 507 550 603 653 705 750 774 805 853 898 Counts 0 0 0 0 0 0 158 886 1120 1185 Voltage (volts) 954 1003 1058 1099 1162 1202 1263 1309 1354 1399 Counts 1192 1238 1258 1206 1221 1256 1295 1371 1507 1712 the plateau is fairly evident from the graph Geiger Plateau 2000 Counts 1500 1000 500 0 400 600 800 1000 Voltage V 7 1200 1400 .Sample calibration data This data was taken by two students in 1995 for one of the Sargent-Welch Geiger-Muller tubes used in the Modern Physics lab course. 26% of all of the data (both sides of the mean). so in time T = N dt we should have a probability of getting M counts of ! µM −µ N M N −M p (1 − p) → ℘(M ) = e M M! where µ = N p. In the experiment we will attempt to verify that the frequency with which the Geiger counter gets X counts per time T is distributed as a random variable. (1 − p) that it will not.4985 0 2πσ 2 In other words. in other words a Poissonian distribution. 2.4 Experiment I.which indicates an operating voltage around 1038 volts. For data that is distributed Normally. If this hypothesis is true. within one standard deviation of the mean you will find 68. and within 2σ of the mean you will find 95. how many data should occur within one standard deviation of the mean? Z σ x2 1 √ e− 2σ2 dx = 0.47725 0 2πσ 2 Z 3σ x2 1 √ e− 2σ2 dx = 0. 8 . Normal distributions We can use nuclear decays to learn a little bit about the Normal distribution. Consider the following set of 104 Geiger counter counts for a Barium sample. assuming that the decay process is random and Normally distributed.34135 0 2πσ 2 Z 2σ x2 1 √ e− 2σ2 dx = 0. demonstrating that nuclear decays are random.45% of the data. in time dt there is probability p that a nucleus in the sample will decay. meaning by a frequency determined by a PDF equal to that of a Normal distribution. each taken over the same time period. mean and standard deviation program */ /* gcc sort. 9 ."%d". /* simple data sorting. exit(0).0.smallest."r"). /* read data file named on command line */ n=0.i++){ if(numbers[i] < smallest) smallest=numbers[i]. for(i=1. &numbers[n]). }while(!feof(fptr)).c -lm */ #include<stdio. smallest. one per line) from a file into an array. } fptr=fopen(argv[1].n. do{ fscanf(fptr.h> int numbers[200].sigma. /* we overshot */ /* get largest.i. if(numbers[i] > largest) largest=numbers[i]. n=n-1. FILE *fptr. do stats */ smallest=numbers[0].sum.i<n.h> #include<stdlib.} mean=0.459 449 430 387 410 396 458 427 422 398 449 405 415 436 459 407 434 418 437 409 363 461 444 435 410 407 421 406 438 402 381 405 434 399 460 449 422 377 389 442 438 445 414 437 377 443 438 423 454 424 447 433 410 411 435 398 421 440 396 444 408 383 374 423 390 430 394 396 406 433 440 417 422 391 430 394 431 420 409 423 438 421 418 389 438 393 431 428 436 433 392 426 434 424 433 423 444 401 414 453 413 433 455 406 We can write a little C program to read this data (as a long list of numbers.h> #include<math. double mean./sort datafile\n"). as well as a largest and smallest value. from which an unbiased estimator for the mean and standard deviation can be gotten. int largest. main(int argc. largest=numbers[0]. n=n+1. char *argv[]){ if(argc!=2){ printf(". i++) sigma=sigma+((double)numbers[i]-mean)*((double)numbers[i]-mean). largest.i++) if(numbers[i]<= mean && numbers[i]>= mean-sigma) sum=sum+1. /* how many are within sigma of mean.mean+sigma).i<n.n.694745 Notice that there are 25 + 42 = 67 of the 104 data within one standard deviation of the mean.sum.i<n. printf("There are %d between mean=%f and mean-sigma=%f\n". we use another short program.sum.846154 21.mean-sigma). If we want to sort these data to create a frequency plot. } This program can be downloaded from the 499 archive.for(i=0.mean. If we run it on our data we obtain gcc -o sort sort. /* print the number of data. 10 . graphing the number of data that fall within a certain lower and upper limit.848591 There are 25 between mean=420. for(i=0. smallest.largest. to left ? */ sum=0. fclose(fptr).846154 and mean-sigma=398. mean and standard deviation */ printf("%d\t%d\t%d\t%f\t%f\n".846154 and mean+sigma=442. for(i=0. this is 64.i++) mean=mean+(double)numbers[i].4%. mean. sigma=sqrt(sigma/(double)n).i<n./sort data2 104 363 461 420. to right ? */ sum=0.sigma).997562 There are 42 between mean=420.i<n. mean=mean/(double)n.c -lm . confirming the notion that the number of nuclear decays per fixed time interval is most likely distributed Normally. suitable for graphing. /* how many are within sigma of mean. for(i=0.smallest. printf("There are %d between mean=%f and mean+sigma=%f\n".i++) if(numbers[i]>= mean && numbers[i]<= mean+sigma) sum=sum+1. sigma=0. mean.0. agreeing quite well with the values computed for a Normal distribution. bins[i]).n.i<n. For example. main(int argc. n=n+1.i++){ which=(int)floor(((float)numbers[i]-(float)smallest)/div). div=((float)largest-(float)smallest)/(float)N. to create histogram plot /* gcc pidgeon. /* pidgeon hole each bit of data */ for(i=0. do{ fscanf(fptr. /* we overshot */ /* zero all bins */ for(i=0."r").h> #include<math. FILE *fptr. /* read data file named on command line */ smallest=atof(argv[2]). if(argc!=5){ printf(".i++) printf("%f\t%d\n". } for(i=0."%d". char *argv[]){ float div.i<N.c -lm 11 .i. n=n-1.c -lm #include<stdio./pidgeon datafile min max number\n").h> */ */ int numbers[200]. our data above could be made into a histogram with 22 divisions between 360 and 470 with gcc -o pidgeon pidgeon.N.smallest. } This we can pass parameters to. n=0.smallest+(float)i*div+div/2. bins[which]=bins[which]+1. int largest. N=atoi(argv[4]). } fptr=fopen(argv[1]. &numbers[n]).i++) bins[i]=0. largest=atof(argv[3]).bins[100]. }while(!feof(fptr)).sum. exit(0).h> #include<stdlib.i<N./* simple data sorting.0.which. xmax=(double)atof(argv[2]). #include<stdio.000000 435.000000 360 470 11 1 3 5 12 12 11 16 24 12 6 2 or piped directly into a graph with .000000 395. You may wish to determine exactly what fraction of the total data falls between xmin and xmax for a Normal distribution of variance σ 2 .xmax. double xmin.000000 375./pidgeon data2 365.x. char *argv[]){ if(argc != 4){ printf(". exit(0).000000 425. } xmin=(double)atof(argv[1]).000000 415. int n./normal xmin xmax sigma\n"). main(int argc.h> #define PI 3..sum.000000 405.sigma.h> #include<stdlib./pidgeon data2 360 470 11 | graph -T X This program is giving the midpoint of each division.000000 385.1415926 double dx. sigma=(double)atof(argv[3]).000000 465. 12 . versus the population of that division. For zero mean this is the total number of data points times the probability ℘= Z xmax xmin √ x2 1 e− 2σ2 dx 2πσ 2 which is computed with the program below.h> #include<math.000000 455.000000 445. The C program below will generate N Normally distributed random numbers with mean mean and standard deviation sigma by the accept-reject algorithm. n=0. float mean. You may wish to simulate the experiment before you actually perform it. max=mean+5. float f( float x. float min. if(argc != 4){ printf(". /* how many to generate */ srand(17). printf("%f\n".test. You can use it to simulate the outcomes of he nuclear counting experiments. } sum=sum*dx/(sigma*sqrt(2.n++){ x=xmin+(double)n*dx. int N. N=atoi(argv[3]).h> #include <math.0. and use the other programs to analyze the resulting data.sigma.0*sigma*sigma)).h> #include <stdlib. sum=sum+exp(-x*x/(2. float sigma).} mean=(float)atof(argv[1]).1415926 int n.mean.m2.0*sigma. main(int argc. char *argv[]){ float m1. sigma=(float)atof(argv[2]). 13 .dx=(xmax-xmin)/10000. to fit your count frequency to a Gaussian Normal distribution rather than the Poissonian example.0.sum).max.h> #define PI 3. crap. /* gcc -o accept_reject accept_reject.0*sigma. /* initialize random number generator */ min=mean-5.c -lm */ /* creates Normal deviates */ #include <stdlib. sum=0./random mean sigma number\n").n<10000. exit(0). You might want to use this program for the next nuclear counting experiment. } You can easily shift the mean of the distribution to the point of your choice.0*PI)). for(n=0. float sigma){ /* Normal distribution */ float y. This data was taken in 1997.2533141/sigma))*exp(-y*y/2)). mean+4*sigma */ m1=min+(max-min)*(float)rand()/(float)RAND_MAX.mean.5 Experiment II.} }while(n<N).do{ /* get a random number between mean-4*sigma. m2=f(m1. } float f( float x. n=n+1. float mean. The Geiger tube voltage has been set at 1040 volts. the better will be your results. return( ((1. Hypothesis testing and goodness of fit Below is a table of the number of counts for 600 five second intervals. The number 400 is a minimum. if(m2>test){ crap=floor(m1). The more counts you make. so that we have time to record our numbers. This requires counting for over two hours. the scaler-timer counts for 5 seconds at a time with 5 second intervals between counting periods. I suggest that you take twice that many. crap). Do it in 30 minute shifts with your lab partners. 14 . test=((1. } 2.sigma).2533141/sigma))*(float)rand()/(float)RAND_MAX. printf("%d\n". y=(x-mean)/sigma. To analyze this data. which sorts the data into 20 equivalence classes and predicts the correct Poissonian parameter (using the unbiased 15 . nuclear decays are random processes. I ran it through binsort in the support software. with a frequency that is a Poisson random variable. We could use a Normal PDF as well.459 422 434 410 434 438 454 421 390 422 438 436 444 422 427 449 398 418 407 399 445 424 440 430 391 421 433 401 425 424 430 449 437 421 460 414 447 396 394 430 418 392 414 423 432 387 405 409 406 449 437 433 444 396 394 389 426 453 423 422 410 415 363 438 422 377 410 408 406 431 438 434 413 432 429 396 436 461 402 377 443 411 383 433 420 393 424 433 390 450 458 459 444 381 389 438 435 374 440 409 431 433 455 455 392 427 407 435 405 442 423 398 423 417 423 428 423 406 421 424 418 453 418 398 408 438 443 412 433 440 417 425 427 403 414 398 401 408 401 408 454 390 437 410 416 408 446 409 425 396 388 423 441 394 414 420 424 396 417 405 429 395 381 438 403 445 393 424 431 419 404 398 423 410 445 430 402 428 407 453 429 405 389 482 369 406 417 426 402 393 456 428 402 407 406 435 433 435 417 450 426 445 443 446 424 393 433 438 424 405 409 428 410 400 408 412 428 435 431 395 419 431 430 428 461 434 433 406 404 403 404 420 425 417 433 392 391 439 431 419 417 405 398 435 407 446 437 458 402 415 432 408 383 374 418 400 387 410 360 385 445 427 439 423 465 434 413 435 416 430 442 437 418 447 456 426 413 396 435 430 433 360 434 428 454 396 408 413 406 406 391 431 437 450 382 413 422 406 437 393 438 387 430 425 415 429 430 393 381 437 479 409 396 412 428 426 423 362 424 411 432 399 432 397 422 424 410 449 415 478 420 416 421 417 417 408 466 436 451 412 433 407 419 460 394 427 415 432 429 410 449 421 381 423 435 402 419 432 453 416 447 404 386 434 445 399 393 461 399 417 434 432 425 429 445 450 402 412 407 414 423 420 404 405 425 385 448 452 435 418 419 436 425 475 418 426 378 469 429 438 453 432 418 368 400 419 439 423 356 428 382 415 422 402 437 444 417 364 448 443 383 435 411 415 415 441 393 394 424 371 399 446 407 433 412 373 437 416 410 432 432 409 468 399 407 425 404 434 399 401 402 401 446 393 388 445 459 369 409 429 441 390 439 436 438 408 449 430 417 447 391 404 412 414 409 424 439 412 394 419 428 408 426 401 442 428 412 425 408 425 434 423 400 444 388 406 376 447 429 427 409 372 416 404 439 419 455 401 411 389 399 431 435 410 393 435 407 439 398 433 454 443 431 426 397 405 409 436 432 414 434 472 458 399 408 396 397 465 423 442 446 421 447 423 452 440 420 412 404 459 438 432 415 422 402 414 398 442 390 393 442 436 392 407 437 429 412 419 428 448 426 386 397 381 449 418 437 384 429 389 418 434 455 452 380 418 433 405 400 420 421 418 420 417 396 423 420 435 414 429 We use this data to test a hypothesis. 5 385.052638 1.984291 69.5 433.5 445.th )2 mj.5 463.5 451.977833 0.397901 14. This is a well known phenomenon.178689 58.5 427.5 421.5 391.090901 10.5 361.th 2 ℘(χ ≥ 458.5 385.013061 1.670648 33.630501 51.065824 35. Notice that when mtheor is small.5 397.5 391.023485 0.5 367.148333 Total count = 600.105783 22.5 379.5 409.5 403. Null measurements are to be avoided.5 469. use the estimators for mean and standard deviation of the last section.041427 0.378958 21.263320 0.5 403.5 463. such as for bins far from the most populated.5 421. Best-fit Poisson parameter is µ = 10.359406) = Z 1 ∞ 458.009999 0.356260 0.000000 .5 433.890918 1.504033 45.th 0.480625 4.5 439.5 451.5 415.5 frequency mj 3 3 6 6 14 22 37 42 62 49 59 64 71 64 31 27 18 12 5 2 theoretical freq.359406 χ2 = 19 j=0 mj.527462 73.5 415.669033 7. the contribution to the χ2 statistic is huge. mj.th )2 = 458.945594 1.5 427.383285 3. Below we have the output.5 379.000915 18.5 457.888281 74.5 373.655743 65.5 457.189890 8.5 367.120103 To fit to a Normal curve.estimator).804898 0.5 445. the relative error in an essentially null measurement can be huge. P (mj −mj.359406 2 16 18 2 Γ( 18 ) 2 18 x x 2 −1 e− 2 dx = 0.206336 8.5 upper bound 361.5 439. number of bins (classes) = 20 Bin j 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 lower bound 355.5 373.387675 2.5 469.460505 0.781345 2.5 397.250773 32.5 475.238331 1.407348 0.5 409.133944 0.th 377.553825 (mj −mj.209332 4. 6 35.2 7.1299) = Z 1 ∞ 30. and we find then that ℘(χ2 ≥ 30.1299 18 2 2 Γ( 18 ) 2 18 x x 2 −1 e− 2 dx = 0.488333 Total count = 600.71 63. number of bins (classes) = 15 17 . If we discard this data.5 28.9 56. we see a better result Best-fit Poisson parameter is µ = 7.036486 If instead we sort into 15 equivalence classes. the result is the reduction of χ2 to χ2 = 30.8 49.7 42.4 21.1299.3 14. and just for a few counts.1 0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 Notice that the first three data bins really throw off the whole goodness of fit criterion. 212238 0.457811 0. mj.Bin j 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 χ2 = lower bound 355.5 451.5 P14 j=0 (mj −mj.5 411.3 77.5 395.503520 51.th 64.5 363.5 38.854127 0.097201 .5 443.5 387.5 443.552675 6.8 29.5 419.5 379.4 9.7 0 0 1 2 3 4 5 6 7 8 18 9 10 11 12 13 14 15 16 (mj −mj.394485 13 13 2 2 Γ( 13 ) 2 x x 2 −1 e− 2 dx = 0.497067 43.880062 82.331652 1.625690 0.5 475.431042 0.797043 2.5 459.957939 82.335745 2.th )2 mj.394485 ℘(χ2 ≥ 88.1 19.5 435.5 467.413470 23.5 435.860687 0.th )2 mj.5 459.000000 97 87.021559 11.5 467.988468 65.th upper bound 363.514170 9.361637 1.481717 0.5 387.5 403.6 67.714187 = 88.472291 1.5 451.5 427.921311 21.5 403.5 frequency mj 5 5 8 19 40 56 77 76 81 97 56 40 25 8 4 theoretical freq.5 379.5 371.297719 34.394485) = Z 1 ∞ 88.5 371.th 0.5 411.651190 1.9 58.5 419.2 48.5 427.332296 68.221978 87.5 395.791868 12.738606 0. 1396 2 13 2 Γ( 13 ) 2 13 x x 2 −1 e− 2 dx = 0. this gives ℘(χ2 ≥ 21. It should be clear that this experiment is sensitive to the size of the data set.In this case discarding the first two bins reduces χ2 to χ2 = 21. 19 . we have a good fit.070014 We read this as saying that if the deviations from the best fit Poissonian of less than 7% are accepted to be insignificance. nuclear decays are random processes. and using the chisquared table generator in the support software. These classes will contribute most to χ2 but will account for the smallest percentage of the actual data.13964. and our hypothesis is verified to be true. the better. The analysis should be performed by regarding the contributions of data classes far from the most populated classes in the proper light.1396) = Z 1 ∞ 21. the larger. Radioactive Barium such as that used in this experiment decays by gamma ray emission. 3. and the right hand side contains the stable isotope. is illustrated below. The scaler-timer simply records decays produced by ionizing radiation emitted by the Barium and detected by the Geiger-Muller tube. Other types of nuclear decay modes are α decay p mX 4 →p−4 m−2 Y +2 He in which an α particle or Helium nucleus is emitted. in particular the one leading to the formation of Radon gas.1 Background The Cesium sample will be used to determine the operating range (voltage) for the GeigerMuller tube. and it’s inverse process n0 → p+ + e− + ν¯e the common decay mode of the neutron. 20 . each decay releasing a gamma ray of 0. electron capture p+ + e − → n0 + ν e in which a neutrino is produced with a neutron and β decay p+ → n0 + e + + ν e a decay mode of the proton in an unstable nucleus producing a positron and neutrino.3 The Half-life of Ba137 The purpose here is to gain more experience with statistics and to learn how to use the Geiger counter.662 M eV 137 m 56 Ba →137 56 Ba + γ in which m denotes a meta-stable isotope of the element. A typical decay chain for radioactive isotopes. beginning with N0 . The time it takes for the sample to be cut in half by radioactive decays is N0 λT 1 = N0 e 2 2 21 .We hypothesize that the probability of decay for a particle within time dt is d℘ = λ dt the rate at which the number of particles in a sample of size N is changing due to decay is then dN = −N λdt which integrates to an exponential decay law N (t) = N0 e−λt for the number of particles left after time t . Scaler-timer. Our Barium sample has an activity of 9 µCi. since we count for ten second periods. The count rate itself exponentially decays along with N . 4. 9µc. The standard measure of radioactivity of a sample is called it’s activity which is the number of decays per second . Barium sample . Cesium Sample. miniature radio-isotope generator.or ln 2 2 λ We can compute the mean lifetime τ¯ by finding the probability that a particle has not decayed by time t. catalog number S-72095-85. the GeigerMarsden experiment was conducted with a 0. Let ∆N be the number of decays detected between t and t + ∆t.7 × 1010 Bq The activity R(t) at a given time is R(t) = −dN = λN0 e−λt = R0 e−λt dt Since the activity is actually the decay rate. To put this into perspective.1 Ci radium sample! 3. Sargent-Welch. 3. Amersham-Searle. Apparatus Geiger tube. the fraction of particles that have not decayed is T1 = f (t) = and so the mean lifetime is N (t) = e−λt N0 1 λ 0 which is what we will compute in the experiment. one decay per second being called a Bq or Becqueral. 2. Sargent-Welch catalog number S-72095-10. 22 . 0. but rather the number of decays. we see that our experiment will actually measure this quantity. During the experiment we will not be able to directly measure the number of remaining nuclei.2 1. Redco mini-generator. N0 − N (t) which also drops with the same exponential rate.1 µc. We use the Ci or curie instead 1 Ci = 3. spaced by twenty second intervals. Since N (t) = N0 e−λt is the number remaining at time t. then τ¯ = Z ∞ te−λt dt = ∆N = N (t) − N (t + ∆t) = N0 (1 − e−λ∆t )e−λt = ∆N0 e−λt the quantity ∆N0 is fixed by the size of the interval ∆t and so we see that the actual number of decays detected will also exponentially drop as the number of particles capable of decay drops. We find a total of 198 counts. and the residue collects on a small disc of blotter paper placed below the generator. The Barium sample is placed under the counter and the experiment begins.3 counts per 10 second interval 23 .The Barium half-life is 2. count for ten. and count the sample for 600 seconds. An acid solution is dripped into the generator which washes the surface of a Cesium/Barium sample.4 Data and analysis Time 0 30 60 90 120 150 180 210 240 270 300 330 360 390 420 450 480 510 540 570 600 Cnts per 10 sec 139 103 87 72 60 66 41 37 41 26 22 18 19 14 11 13 11 10 6 8 5 We now need to correct for background radiation. count for ten. In other words. Barium is produced from the Cesium by a beta decay 137 55 Cs − →136 ¯e 56 Ba + e + ν The Geiger counter operating voltage is now determined by counting the other radioactive source. rest for twenty and so on. rest for twenty.3 Procedure The first step is to collect a Barium sample from the mini radio-isotope generator. or 3. 3. The Cesium isotope used has a 27 year half-life and so provides a constant rate of decays for the counter.6 min. Wait for one hour.3. Cesium. We will count for ten second intervals separated by twenty second periods. 80517 3.32435 2.7 6.91833 4.7 4.91045 9.993252 2.83406 2.27109 2.can be attributed to background radiation.60217 9."&".7 σCnts0 ln |Cnts0 | 11.awk { print $1.7 37.03777 7. We can compute the error in the log of the number of counts ∆ ln Cnts0 = σCnts0 1 ∆Cnts0 = =√ 0 0 Cnts Cnts Cnts0 and produce a table of file of data suitable for error-bar graphing with the awk script 24 .11448 2.77489 2.22975 7.3702 2.12236 4.7 22.62966 5.14003 3. √ σCnts0 = Cnts0 is the standard deviation in the number of counts."&".$2.04122 3.13836 6.7 9.92852 3.42724 8. We now correct the data for this and work it up using the awk script #analyze."\\\".27213 2.75366 3.54756 1.7 1.62966 4.3).04122 2.7 7.90211 1.7 68.7 99.3).7 83.sqrt($2-3.7 33.64317 0.5175 6."&".98499 4.3.log($2-3.7 14.76445 3.96232 2.7 18.52994 4."&".14003 3.$2-3.7 7.530628 with Cnts0 = Cnts − Cntsback is the corrected number of counts per 10 second interval.649 4.28855 4.77489 2."\\hline"} This results in the table below Time 0 30 60 90 120 150 180 210 240 270 300 330 360 390 420 450 480 510 540 570 600 Cnts per 10 sec 139 103 87 72 60 66 41 37 41 26 22 18 19 14 11 13 11 10 6 8 5 Cnts’ 135.7 15.7 10.7 56.58844 1.14877 4.68785 3.7 37.7 2.7 62.16795 1.30384 0. 04122 2.awk { print $1. You could use gnuplot which has error bar graphing.386334 0.92852 2.42724 4.90211 0.10015 0.132803 0.91045 4.209888 0.085844 0.12236 2.04122 1.766965 The graph of which is below.62966 3.log($2-3.26082 0. 1.231249 0.30571 0.360375 0.126289 0.162866 0." ".13836 3. 25 .3)} Time 0 30 60 90 120 150 180 210 240 270 300 330 360 390 420 450 480 510 540 570 600 ln |Cnts0 | 4.68785 2. created by lg in the support software.60217 4.03777 4.22975 4.120648 0.993252 1.321081 0.360375 0.62966 3.530628 σln |Cnts0 | 0.162866 0.109304 0.75366 2.54756 0.5175 3. " ".461266 0.3).3702 2.608581 0.17226 0.252377 0.0/sqrt($2-3.27213 2.#analyse2. 28 sec |m + σm | |m| |m| and ln 2 ln 2 σm ≈ (1 − + · · ·) = 93. It is questionable scientific procedure to discard data.001414.8 1.005963 1s which places the half-life between the limits ln 2 ln 2 σm ≈ (1 + + · · ·) = 152.6 4.8 3.4 3 2. The linear regression program in the appendix gives us a slope for these 17 points of y = 4. We can run a linear regression on the data . 2 σm = 0.005963 ± 0.6 2.005963x.811657 + −0.114469.4 1 0 51 102 153 204 255 306 357 408 459 510 With log of corrected counts of the vertical axis and time in seconds on the horizontal axis.000002 which results in m = −0. χ2 = 0.5 4. but first we will discard the three data points indicated in the table with an asterisk.2 1.2 3. These arepoints with count totals so close to the background level that they are most probably overwhelmed by the fluctuations in the background count rate. unless there is very good reason.95 sec |m − σm | |m| |m| 26 . τ¯1 = − 2 ln 2 = 116 s −0. 6 min = 156 sec value.which is in fair agreement with the accepted 2. In this way we will have enough counts at subsequent times that background fluctuations do not overwhelm our counts. the Nuclear Shell Model The actual potential felt by nucleons in a nucleus is complex and the Schroedinger equation is impossible to solve analytically. L and so we decompose the wavefunction into a radial part and a spherical harmonic ψ = R(r) Y`. Lz ] = 0 [H. It is very useful for predicting stable nuclei configurations. three dimensional oscillator in polar coordinates. 3. When we perform this experiment we should be sure that enough isotope is collected that we have at least 600 or more counts initially.m (θ. Since the potential is spherically symmetric we find ~ 2 ] = [H. Begin with the nonrelativistic.5 Background material. However for deeply bound nucleons the potential looks a lot like a three dimensional oscillator and the nuclear shell model is based on this potential. φ) and the radial equation becomes 2m h ¯ 2 `(` + 1) mω 2 r 2 d2 2 d + 2 (E − ))R(r) = 0 ( 2 = − dr r dr 2mr 2 2 h ¯ Perform the variable change ξ= mωr 2 h ¯2 s d mω q d ξ =2 dr dξ h ¯2 and we find d2 ξ `(` + 1) ξ 3 d +( − − ))R(ξ) = 0 + 2 dξ 2 dr 2¯ hω 4ξ 4 We now impose the boundary conditions on the wavefunction (ξ R(0) = 0 R(∞) = 0 by redefining the wavefunction l ξ R(ξ) = ξ 2 e− 2 u(ξ) and we find that 27 . 1. x) = ∞ X (a)j xj . M − 2. 2. M − 4. The plot of energy versus angular momentum forms straight lines called Regge trajectories.· · ·. θ. x) + (c − x)f 0 (a. c.m (r. 2. D(M ) = You will often see this written in the literature in terms of the oscillator radial quantum number n as E ` 1 n= − + 2¯ hω 2 4 with solutions that can be written in terms of the Laguerre polynomial. c. c. 3. ` can be M . We know that a series solution exists f (a. · · · is (M + 1)(M + 2) 2 For a fixed value of M . · · ·. The spectrum that we obtain is −a = 3 E =h ¯ ω(2N + ` + ) 2 Each ` value has 2`+1 degenerate m` values and a little arithmetic shows that the degeneracy of a level of energy 3 E =h ¯ ω(M + ) 2 with M = 0. φ) = An. φ) with r ρ= . j=0 (c)j j! (a)j = a(a + 1) · · · (a + j − 1) which will truncate at a polynomial (satisfying the condition at ∞) if E 3 ` − − =N 2¯ hω 4 2 where N = 0.m (θ. mω ξ = ρ2 . 1. which is a hypergeometric function 1 2 `+ 1 Ψn. however ` must remain positive. c.(ξ d2 3 d E 3 ` + ( − ` − ξ) + ( − − ))u = 0 dξ 2 2 dξ 2¯ hω 4 2 and again we recognize the same confluent hypergeometric equation x f 00 (a. r0 r0 = s 28 h ¯ . x) = 0 that has appeared in every potential problem that we have studied in Physics 441.` e− 2 ρ Ln−12 (ρ2 ) Y`.`. x) − af (a. 19 A 3 − 0.44 This is pretty useful for computing γ ray energies for nuclei that decay through γ emission. Z ∞ 0 m xm e−x Lm k (x) Lk 0 (x) dx = Γ(m + k + 1) δk. To do this we compute Z Ψ∗n. and results in 1 hr 2 i = (2n + ` − ) r02 2 and from the rather extensive experimental data. m` > can contain two protons and two neutrons.` = v u u t 2n+`+1 (n − 1)! √ (2n + 2` − 1)!! π r03 We now try to find the oscillator frequency ω that best fits the experimental data as far as the size of the nucleus and the binding energy of a nucleon in the nucleus is concerned.7 M eV 1 h ¯ω = 1 . rr = 1.`. in the nuclear shell model we assume that nuclei are stablest when they have nucleons filling each shell . one spin up. This computation is a good excerise in quantum mechanics and the various recursion relations satisfied by the Laguerre polynomials. `.k0 k! which will allow one to show that the normalization factor for the wavefunctions is An. we have the result that the nucleus containing A nucleons subjects its nucleons to a potential pretty well described by a harmonic oscillator with q 34.44 f ermi A 3 − 0.one spin down of each.m r 2 Ψn.m d3 x = hr 2 i and fit this to the experimental radius-squared of a nucleus containing a total of A nucleons.and the Laguerre polynomial satisfies the recursion Lm k (x) = 1 x −m dk k+m −x e x (x e ) k! dxk These functions are orthonormal on [0. ∞] with respect to kernel xm e−x . Each state |N.`. Analogous to noble gasses with filled atomic shells. This gives rise to stable nuclei of atomic number (Z) 2=Z 2+6= 8=Z 2 + 6 + 2 + 10 = 20 = Z 29 . `. mj (r. You hopefully remember that that means Jz Vj. altering the potential to 2α ~ 1 · ~s V (r) = −V0 + mω 2 r 2 − 2 L 2 h ¯ The eigenstates of the full Hamiltonian become 1 2 `+ 1 Ψn. But and so ~ 2 Vj.`. 20 nuclei are known to be very stable.mj (θ. Indeed the Z = 2.mj j ~ 2 = (L) ~ 2 + 2L ~ · ~s + ~s2 (J) ~ · ~s = j(j + 1) − `(` + 1) − s(s + 1) L 2 and by the rules of angular momentum addition. 2 or j =`− 1 2 We find then that the energy En.2 + 6 + 2 + 10 + 6 + 14 = 40 = Z These are called Magic Number nuclei. After that the nucleon spin-orbit coupling disrupts the energy levels and breaks the degeneracy.`+ 1 .`. until we reach Z protons and A − Z neutrons. We define the term symbol of a single nucleon in a way very similar to that of an electron in an atom. φ) = An.` of these states will be 1 hω − V0 − α` En. since s(s + 1) = 21 ( 12 + 1).`.j.` e− 2 ρ Ln−12 (ρ2 ) Vj=`± 1 . we begin adding nucleons to a shell model manifold of single particle states. j.m = h (J) ¯ 2 j(j + 1) Vj.`. 8.mj and from quantum theory. θ.`. φ) 2 in which the angular part of the wavefunction is an eigenstate maximal set of commuting operators derivable from the total angular momentum operator ~ + ~s = L ~ + 1h J~ = L ¯ ~σ 2 containing contributions from nucleon orbital and spin angular momenta. 1 j =`+ .mj = h ¯ mj Vj. and alters the magic number pattern. Even better predictions of the magic numbers A of stable nuclei come from including the effects of the spin-orbit interaction beteen nucleons.` = (2n + ` − )¯ 2 2 30 . with nomenclature (N + 1) Lj = n Lj and the nuclei of various elements are constructed by a nuclear Aufbau process. 2 1 ¯ ω − V0 E1.` = (2n + ` − )¯ 2 2 This results in the single-nucleon states shown below. which is in very good agreement with experimentally established stable nuclei-nuclei with filled shell-model shells. α1.0 = h 2 2 and the table below together with the formula for ω can be used to determine nuclear energy level spacings.2 = 1.and 1 hω − V0 + α(` + 1) En.4 = 0. 28 particles 2 1 2p 3 3¯ hω − α 4 2 2 1 2p 1 3¯ hω + 2α 2 2 1 3 1f 5 3¯ hω + 4α 6 2 1 4 1g 9 4¯ hω − 4α 10 2 fifth magic filled shell.j.3 = 0.1 = 0. and the frequencies of γ rays emitted when a nucleon changes levels. but is small. 1 . 8 particles 1 2 1d 5 2¯ hω − 2α 6 2 2 0 2s 1 2¯ hω 2 2 1 2 1d 3 2¯ hω + 3α 4 2 third magic filled shell. α2. 20 particles 1 3 1f 7 3¯ hω − 3α 8 2 fourth magic filled shell.1 = 0. ` State En. α1. α1. experimental data shows that α depends on `.2 = 0. The state of lowest energy in the shell model is n Lj = 1s 1 .15.` − E1. αn. 50 particles n 1 The superstable magic filled shells are well separated in energy from the next highest set of (empty) one particle states.19 and it continues to decrease for both increasing n and `.76 α2.`− 1 .93.58. Furthermore. α3. An energy level diagram similar to that used to perform the Aufbau process for atomic states is illustrated below.68.` .0 Nj 2 0 1s 1 0 2 2 first magic filled shell. 1 . for example (in MeV). 2 particles 1 1 1p 3 h ¯ω − α 4 2 1 1 1p 1 h ¯ ω + 2α 2 2 second magic filled shell. 31 . Such an isotope will have a filled shell for both types of nucleons. 32 .Remember that we fill in neutrons and protons independently. Example Apply the nuclear Aufbau process to build up the most stable Helium isotope. Z-odd. even-p.2 M eV √ A By pairing we mean. The pairing is with respect to mj . N = A − Z odd. A filled shell has no contribution to the nuclear spin. odd-even nuclei. This is also called odd-odd. Predictions. This means that we need to distinguish between three nuclear types. so we expect pairing between |n. N odd or Z odd. this pairing amounts to the combination of all angular momenta of the shell’s nucleons to j = 0. to describe real. |j − j| = 0. so all protons are paired and all neutrons are paired. −mj i. 2j − 1. Remember that two particles of angular momentum j can have two-particle wavefunctions of angular momentum 2j. This conforms to experimental evidence. · · · . N = A − Z even.1d 5/2 1p 1/2 1p 3/2 1s 1/2 -1/2 1/2 -1/2 protons 1/2 neutrons In order to use the shell model. which describes only single nucleon states. j. multinucleon nuclei. so all protons but one are paired and all neutrons but one are paired. For a filled shell. the nuclear spin of odd-even nuclei is determined by 33 . 2j − 2. `. even-n. This pairing increases the binding energy of the resultant state. j. A-even. we use the Aufbau process to add nucleons to the shell model states computed with the appropriate A. The shell model can be used to determine properties of the entire multinucleon nucleus. Z even. and further take into acount the fact that paired. which are going to be largely determined for odd-even nuclei by the last unpaired nucleon. N even. The spin of an even-even nucleus in the ground state should be zero. `. or Z even. the two combine angular momenta to 0. Due to cancellations by the pairing. or Z odd. like nucleons have an additional pairing energy of Epairing = 2 11. mj i and |n. Several nucleons in an incomplete shell often do combine their angular momenta to produce a different value J 6= j. This means that a nucleus will have the quantum numbers of holes in filled shells. Suppose that there are k ≤ 2j + 1 nucleons in a shell. they may produce a state whose label is (n(`)j )kJ and a perfect example is the Sodium isotope used in several of our experiments 22 N e21 10 . Since in a filled shell all nuclear spins 2 are paired. and therefore so does this nucleus. a consequence of the nucleon urge to pair. this hole has spin one half. The N715 does experimentally 34 . and superstable. N a11 Z 1 3 5 9 11 19 19 25 27 27 A−Z 0 4 6 10 12 20 22 30 30 32 (1d 5 )33 are in state 2 2 Some odd-even nuclei A Name J Last unpaired nucleon 1 1 H 1s 1 2 2 3 3 7 Li 1p 2 2 3 11 B 1p 3 2 2 1 19 F 2s 1 2 2 3 23 Na 1d 5 2 2 3 39 K 1d 3 2 2 3 41 K 1p 3 2 2 5 55 Mn 1f 7 2 2 7 57 Co 1f 7 2 2 7 59 Co 1f 7 2 2 Example O816 is doubly magic. We define the j value of a filled shell to be zero.the wavefunction of the last nucleon. 1d 5/2 1p 1/2 -1/2 1/2 -1/2 1/2 1p 3/2 -3/2 -1/2 1/2 3/2 -3/2 -1/2 1/2 3/2 1s 1/2 -1/2 1/2 protons -1/2 1/2 neutrons O715 is doubly magic except for a proton hole in 1P 1 . the ground state nuclear spin being the angular momentum j of the last unpaired nucleon. The total electrostatic potential energy due to a nuclear charge Z is Ucoul 1 = 4π 2 Z r0 0 Zr 2 1 r ( ) dr + 4π 3 4π0 r0 2 2 Z ∞ r0 r2 ( Z 3 Z2 2 ) dr = 4π0 r 2 5 4π0 r0 in which r0 is the mean nuclear radius.21 F ) A 3 A3 Ucoul = which evaluates to 30. in order to achieve pairing. Nuclear spin and parity actually plays a very imporant role in the spectra of diatomic molecules. Consider a N a11 with an unpaired proton and neutron. is for the unpaired proton to inverse-β decay β + ν γ 1d 5/2 1d 5/2 -5/2 -3/2 -1/2 -5/2 -3/2 1p 1/2 -5/2 -3/2 -1/2 1/2 1p 1/2 -1/2 1/2 -1/2 -1/2 1/2 1p 3/2 1/2 -1/2 1/2 1p 3/2 -3/2 -1/2 1/2 3/2 -3/2 -1/2 1/2 -3/2 -1/2 3/2 1s 1/2 1/2 3/2 -3/2 -1/2 1/2 1s 1/2 -1/2 1/2 protons -1/2 1/2 -1/2 neutrons 1/2 protons 35 -1/2 1/2 neutrons 3/2 . Using the shell model value we see that this is 3 0. a fact verified experimentally. The nucleus O817 is doubly magic except for an extra nucleon in the 1D 5 and so 2 the shell model predicts a nuclear spin j = 25 again in agreement with experiment. This will be evident even in the thermodynamic functions for gases of such molecules.714 Z 2 M eV Z2 1 = 1 5 4π0 (1. Remember that the neutrons are bound a little tighter than the protons due the the raising of the proton energy caused by the Coulombic repulsion of like charged particles. The most likely decay therefore. We can use the concept of pairing energy to understand β decay processes of some of the 22 nuclei used in this course.83 M eV for N a22 11 .have j = 12 . and a filled p shell with each nucleon having ` = 1 and so has even parity. and the parity of the wavefunction will determine which rotational modes exist for the molecule. Nuclear parity under inversion of the coordinate system is given by P = (−1)( for example the nucleus P (nucleon ` values) O816 has a filled s shell . each nucleon having ` = 0. such as N a22 11 . This line of reasoning is probably qualitatively correct. and in the valence shell there is one unpaired proton and one unpaired neutron with jp = jn = 25 . which now quickly pair. A useful source of information is Nuclear Forces by Gernot Eder. emitting a positron that quickly captures an electron from the innermost shell (K shell) of the electron cloud. it be 3±1. leaving a spin-0 nucleus and shedding the excess energy as another γ. The decay leaves a Neon nucleus with an even number of protons and neutrons. 1968. 4 Determination of the Speed of Light At one time the speed of light was thought to be infinite. A correct determination of the speed of light is a crucial experiment in modern physics and can be performed with relatively simple apparatus. Recall that the total nuclear spin of the Sodium N a22 11 was 3 before the decay. and so the nuclear spin is one of the spins in the decomposition 5 5 ⊗ =5⊕4⊕3⊕2⊕1⊕0 2 2 Conservation of angular momentum requires that instantly after the decay. Transformer 120 V AC Rotating mirror Photoelectric tube Fixed mirror Beam splitter Meter stick scale 120 V AC Frequency meter 120 V AC Power supply Laser 36 . the resulting annihilation produces two γ’s at 0.511 M eV that are not shown in the figure.A proton first decays to a neutron. in fact it is experimentally found to be 2. but I have not been able to compute the energy of the last γ-emission correctly from the shell model. Remember that filled shells have nucleon spins paired to zero. MIT Press. Beam splitter. Magnifier lens. Fluke 1900A multi-counter. 7. At an earlier time. and the number of pulses per second is the inverse of the rotation frequency of the rotating mirror.Apparatus 1. 8. 5. Power supply. and the beam (very finely dotted beam) from the laser was directed by the rotating mirror towards the distantsecond mirror. Laser. Spectra Physics Stabilite Helium-Neon laser model #120. 37 . In this case the emitted beam from the laser and reflected beam from the splitter to the meter stick scale have the same path length. it reflects the laser beam into the photo-tube at some point. Spectra Physics laser exciter model #249. The apparatus is set up with about 5 meters between focusing lens and laser. Transformer. The current position of the mirror is position 2. The distance from the meter stick to the beam splitter is the same as from beam splitter to rotating mirror. Consider the figure below. and the returning beam reflected from the splitter will focus to a point more or less on the scale. Frequency meter. This may require that the beam sprayed into the tube be out of line with the beam reflected back into the beam splitter to avoid interference of the two beams. Leybold 476-41. Do Not Look Into The Beam! 3. The rotating mirror is shown below. A glass slide. 4. the mirror was at position 1. one bounces off of the rotating mirror and heads directly to the scale. 6. Rotating mirror. These two beams arrive at the scale at the same time. 9. This is a powerful laser. The Photoelectric tube must be set up in such a position that each time the rotating mirror undergoes a single rotation. showing two beams arriving at the scale. Now it is returning and hits the mirror in it’s current position but does so at an angle such that it is not directed back along it’s original path. Variac type W5M73. Flat mirror and stand. by an angle. Photoelectric tube. 2. but were emitted by the laser at different times separated by the time it takes the light to travel back and forth between rotating and fixed mirrors. θ = ωdt. and tan 2θ = so that c= dt = S D 4π L D ν S Data In this run we measured d five times and obtained d = 53. d=distance from laser to beam splitter=distance from beam splitter to scale θ= change in position of mirror ν=rotation frequency of mirror in Hz ω= natural rotation frequency of mirror.position 1 position 2 θ L 2θ S S=distance between beam traces on scale L=distance between fixed and rotating mirrors D=distance from rotating mirror to laser.9 ± 0.1 ± 0.5cm 38 2L c .03cm and measured D five times finding D = 309. We have the basic relations ω = 2πν. 0cm When the separation S = 0.0 ± 3.2 cm s 8m which is not far off of the accepted value of c = 2.9 cm)(502 Hz) = 3.2 cm it was found that the rotating mirror had frequency 502 Hz and so m 4π(3111 cm)(309. Things to watch out for.and finally L five times obtaining L = 3111. the mirror is two-sided! 39 .041 × 108 c= 0. Calculate ν correctly.9979 × 10 s . This will cause the cell to expand and so the air density and pressure inside of the cell will change.1 Background Sound is a longitudinal compression wave in a material medium such as a gas or solid. P(x+dx) P(x) A A x x+dx x+η x+dx+d η + η Let passage of the wave through the gas cause the left wall to displace by an amount η to the right and the right wall of the layer displace by η + dη.5 Determination of Cp Cv for a gas The specific heat ratio for a gas is known as the adiabatic constant and can be measured by studying a typical adiabatic process such as sound wave propagation in the laboratory. 5. The new cell volume is V0 + ∆V = A(dx + dη) and so the volume change is ∆V = Adη and we find that 40 . We will first derive the wave equation in order to find the velocity of sound and the relation between the displacement and pressure waves. Consider a layer of molecules of gas of cross sectional area A and thickness dx. The old cell volume was V0 = A dx and the mass of the gas in the cell was and still is dm = ρ0 A dx in which ρ0 is the density of the gas at equilibrium. In that case P0 V0γ = (P0 + ∆P ) (V0 + ∆V )γ in which γ is the ratio of specific heats γ= Cp Cv We can expand the right hand side using the Binomial theorem (1 + δx)n = 1 + nδx + · · · for small x. P (x). after all we are dealing with the expansion of an ideal gas presumably. which in our case becomes (V0 + δV )γ = V0γ (1 + δV γ ) V0 ∆V + · · ·) V0 and if we again neglect the product ∆V δP we find that = V0γ (1 + γ 41 . and no heat is exchanged between adjacent gas cells and so propagation of sound is adiabatic. We find that the net force to the right is P (x + dx) = P (x) + Fx = A(P (x) − P (x + dx)) = −Adx ∂P ∂x Since the cell has acceleration ax = d2 η dx2 the equation Fx = m ax reads ρ0 dxA d2 η ∂P = −Adx 2 dx ∂x or d2 η 1 ∂P = − dx2 ρ0 ∂x This is the acoustic equation. the pressure on the left wall of the cell.dη ∆V = V0 dx The force causing these displacements is supplied by pressure gradients. We need to establish the thermodynamics of the process. pushes to the right and that on the right wall ∂P dx + · · · ∂x pushes to the left. Sound propagation is a fast process. A full treatment of the specific heats CP and CV for polyatomic gasses is given at the end of this experiment. 2. 4. 5. Thermometer Gas Out Gas In Speaker Kundt’s tube Moveable Piston Rod Microphone Meter stick or scale Audio Oscillator 5. Apparatus Kundts tube apparatus. Such a phase relation 42 . CO2 . 3. High pressure gas cylinders of Ar. Audio oscillator and speaker. This apparatus is essentially ”canned” and lives in the physical chemistry laboratory. N2 .3 Oscilloscope Procedure An oscilloscope will be used to monitor the phase relationship between the input sound signal into the gas in the tube and the signal picked up by a microphone.∆P = −γ ∂η ∆V P0 = −γ P0 V0 ∂x and the total gas pressure is ∂η ∂x Insertion of this into the acoustic equation gives all the same results as before except now we find that sound waves travel (adiabatically) at speed P (x) = P0 + ∆P = P0 − γP0 v= s γP0 = ρ0 s γR T M where M is the molecular weight in grams per mole.2 1. Oscilloscope. Thermometer. 5. reduce the flow rate to a low value. Before data is ready to be collected. The piston should be positioned near the end of the tube as seen in the figure. look like the following This will require that the apparatus be calibrated by holding a tuning fork near the microphone with the speaker disconnected from the scope input. The apparatus needs to be calibrated at several frequencies around 1 to 2 kilohertz. t) = A0 sin 2π(x + λ2 ) 2πx = −A0 sin λ λ eliminating x we should expect the curve A 0 y A0 or a straight line to be displayed on the scope. 1 kHz for Nitrogen and Carbon dioxide. and slowly move the piston towards the speaker end of the tube. which is not sealed. Set the audio oscillator on the desired frequency. t) = A sin 2πx λ while the microphone receives y 0 (x. y= 43 . to prevent air from diffusing into the tube.can be displayed as a Lissojous figure by displaying one signal on the horizontal input. 2 kHz for He or Ar. Adjust the gain on the scope until both input and microphone signals are at the same magnitude. These runs need to be repeated and statistical analysis performed. The program used to create these figures is included at the end of this experiment. When a half wave is standing in the tube the waveform being input will be y(x. Typical Lissojous figures for example with y 0 (t) = Ay sin(ωy t − φ) y(t) = Ax sin(ωt). and the other on the vertical input to the scope. and the tube flushed for about 10 minutes with the gas to be studied. but not zero. 50 13.8 ± 1.35 13.54 ± 0.45 16. cm 2 v.0C = 298K and at frequency ν = 1000.0 270.65 16.97 44.01 39.20 13.15 13.20 17.303 ± 0.10 13.0Hz Using the ideal gas constant R = 8.15 13.08 16.10 ± 0.01 44 .673 ± 0.31 J moleo K we obtain Quantity λ .55 16.4 Data and analysis This data was taken by a student in spring 1997.05 All measurements were taken at temperature T = 25.0 ± 3. λ 2 in cm Air CO2 Ar 17. mole p γ=C Cv Air CO2 Ar 17.6 322.5.948 1.05 346. ms gm M.05 17.40 13.54 ± 0.024 1.0 28.10 17.30 ± 0.25 13.10 17.10 ± 0.400 ± 0.0 ± 1.45 16. Trial 1 2 3 4 5 6 Ave.30 ± 0.05 17.015 1.08 16.60 16.55 16.10 17. The exact values are as follows. Diatomic molecules have one vibrational, two rotational and three translational degrees of freedom., monatomics have only three translational, and Carbon dioxide has vibrational modes, you should read the next section thoroughly in order to see where the theoretical values come from. Quantity γexp γtheory % difference 5.5 Air CO2 Ar 1.400 ± 0.024 1.303 ± 0.015 1.673 ± 0.01 1.400 1.285 1.667 0.0% 1.4% 0.38% Molecular spectra and the equipartition theorem ) and CV = ( ∂U ) in a quantum mechanically correct The precise form of both CP = ( ∂U ∂T P ∂T V determination are both temperature dependent, both look like “steps” or plateaus, indicating that certain molecular motion modes are excited at various temperature regimes. What modes are we talking about? There are three types of molecular motions; translations and rotations (purely kinetic), and vibrations (possessing both kinetic and potential energy). We can compute the specific heat of a molecule with classical formalism for a simple illustrative case such as the linear molecule CO2 . Consider then the lagrangean for the molecule fixing all atoms to lie on a line; 1 1 1 1 1 L = mC z˙ 2 + mO (z˙ − x˙ 1 )2 + mC (z˙ + x˙ 2 )2 − k(x1 − `)2 − k(x2 − `)2 2 2 2 2 2 in which z fixes the position of the Carbon in space, and xi is the distance between each Oxygen and the Carbon. The equations of motion are d ∂L ∂L ( )= dt ∂ z˙ ∂z or mC z¨ + 2mO z¨ − mO x¨1 + mO x¨2 = 0 d ∂L ∂L ( )= dt ∂ x˙ 1 ∂x1 or mO x¨1 − mO z¨ = −kx1 and finally ∂L d ∂L ( )= dt ∂ x˙ 2 ∂x2 or mO x¨1 + mO z¨ = −kx2 Following the usual T and V matrix methods for normal modes, we can write this as       mC + 2mO −mO mO z¨ 0 0 0 z      −mO mO 0   x¨1  =  0 −k 0   x1    mO 0 mO x¨2 0 0 −k x2 45 We assume that each coordinate is an amplitude times eiωt , which results in a secular equation for the amplitudes and frequencies −(mC + 2mO )ω 2 mO ω 2 −mO ω 2 Az    mO ω 2 −mO ω 2 + k 0   A x1  = 0  −mO ω 2 0 −mO ω 2 + k A x2    The frequencies are determined by the vanishing of the determinant 0 = −(mC + 2mO )ω 2 (−mO ω 2 + k)2 − (−mO ω 2 + k)(mO ω 2 )2 This gives us three frequencies; ω02 = 0, k , mO ω12 = ω22 = (mC + 2mO ) k mC + m O mO In terms of these normal modes, we could perform a coordinate transformation      z O1,1 O1,2 O1,3 ξ0       ξ1  =  O2,1 O2,2 O2,3   x1  x2 O3,1 O3,2 O3,3 ξ2 in terms of which the classical lagrangian would be 1 1 1 1 1 L0 = ξ˙02 + ξ˙12 − ω12 ξ12 + ξ˙22 − ω22 ξ22 2 2 2 2 2 The classical Hamiltonian will be 1 1 1 1 1 H0 = p20 + p21 + ω12 ξ12 + p22 + ω22 ξ22 2 2 2 2 2 which results in the classical partition function Z= Z dp0 dp1 dp2 dξ0 dξ1 dξ2 −βH0 e , h3 β= 1 kT It is a good exercise to compute this function by doing the integrals, each is from −∞ to ∞, to compute the average energy of such a molecule in contact with a heat reservoir at temperature T ∂ ln Z 5 U¯ = − = kT ∂β 2 and the specific heat at constant volume CV = ( 5 ∂ U¯ )V = k ∂T 2 In a one dimensional world, we find that our gas of such molecules will have γ= CV + k CP = = 1.4 CV CV 46 The Equipartition theorem is essentially just a rule of thumb gotten by inspecting the Hamiltonian written in normal mode coordinates; CV gets a contribution of 12 k from each quadratic term in H0 . In this case there are five terms. In three dimensions, we will need three coordinates (x, y, z) to specify the position of the Carbon in space, the two bond distances, and two Euler angles θ and φ to specify the orientation of the molecule in space. In that case the classical Hamiltonian will be (in normal mode coordinates) Pφ2 1 ~2 1 1 2 1 2 2 1 2 1 2 2 2 H = P + (Pθ + 2 ) + p 1 + ω1 ξ 1 + p 2 + ω2 ξ 2 2 2I 2 2 2 2 sin θ 0 Counting quadratic degrees of freedom, we find nine, and so CV = 92 k, and CP = CV + k making the adiabatic constant 1.2222. In the real world we find that stiff molecular vibrational modes are generally not excited unless the temperature is quite high. This can only be seen from a quantum mechanical treatment of the problem. If we toss out the four vibrational degrees of freedom, we are left with five excitable molecular modes at low (room) temperature. This gives us γ = CCVP = 75 = 1.4 for CO2 . The full quantum treatment of only the vibrational modes of this molecule will give us an average energy h ¯ ω1 h ¯ ω2 ¯ = 5 kT + U h ¯ ω1 + h ¯ ω2 2 2 sinh 2kT 2 sinh 2kT since the quantum Hamiltonian will be P2 1 1 1 1 H0 = P~ 2 + (Pθ2 + φ2 ) + h ¯ ω1 (n1 + ) + h ¯ ω2 (n2 + ) 2 2I 2 2 sin θ and the partition function for the quantized oscillation terms will be ∞ X 1 e−β¯hωi (ni + 2 ) ni =0 (sums rather than integrals). The rotational modes also will have a temperature dependence, an issue that we will not go into here. The corresponding specific heat will be h ¯ ω1 h ¯ ω2 5 h ¯ 2 ω22 cosh 2kT h ¯ 2 ω12 cosh 2kT CV = k + h ¯ ω1 + h ¯ ω2 2 4kT 2 sinh2 2kT 4kT 2 sinh2 2kT This has the adversised plateau structure, here plotted for ω2 = 6.0 · ω1 ; 47 5 4 CV/k 3 2 1 0 0 2 4 6 8 10 2πkT/h You can see that the specific heat is roughly constant at some multiple of k, as classically predicted, but only in certain temperature regimes. This is a clear indication that at sufficiently low temperatures, the oscillatory modes are not excited and cannot absorb heat. It turns out that CO2 has an additional oscillatory mode, a flexing out of the linear geometry, that has a much lower frequency (because of weal restoring forces) than either bond-stretching mode. This comes out of a group theory treatment of molecular vibrations. If we work at a low enough temperature that all three translational modes, both rotational, and the flexing mode (Potential and kinetic degrees) are excited, the classical prediction for CO2 will be CV = 27 k, and the adiabatic constant is then γ = 97 = 1.2857, which is reported as the theoretical value in the writeup. You should verify that for Ar, a monatomic, CV = 32 k and γ = 53 = 1.666, and for N2 or O2 , if we neglect the vibrational modes of the extremely rigid double bonds, CV = 25 k and γ = 57 = 1.4. 48 The high pressure gauge should drop to zero. set the regulator to zero. This amount will be adequate for the experiment. and never use tools to force open or close a valve. Open the shutoff valve. Now turn the outlet valve. until the low pressure gauge rises slightly above zero. Slowly turn the regular handle clockwise. the regulator handle and outlet valve. The high pressure gauge will now read tank pressure.c -L/usr/X11R6/lib -lplot -lXaw -lXmu -lXt -lSM 49 -lICE -lXext -lX11 -lm .7 Lissojous program /* Draw lissojou figures for illustration */ /* gcc lissojou. first turn off the outlet valve (close it). When you are done. Set the regulator to zero flow by turning the regulator handle counter-clockwise until you feel resistance diminish.5. 5. If it is already set to zero flow it will turn freely. Even if the low pressure gauge reading drops there is still gas flow. then close the shutoff valve. Be sure the outlet valve is closed firmly but never forced.6 Safe Handling of High Pressure Gas Cylinders High pressure gauge Low pressure gauge Shutoff valve Regulator Outlet valve Regulator handle Gas hose Gas Cylinder Locate the shutoff valve on top of the cylinder. Open the outlet valve. Gas cylinders should always be secured to the workbench.then the regulator slightly. Once again close the regulator and outlet valve. screwing it in. allowing gas trapped in the regulator to escape. pl_filltype(0).h> #include<plot. "Couldn’t open Plotter\n").0).*labels. char Wxl[14]="\\*w\\sbx\\eb=". pl_handle = pl_newpl (cmdptr.h> #include<stdlib. stderr). exit(0). cmdptr=argv[6].Ax.0).0*PI/(1000. char buffer[6]. pl_erase(). return 1. pl_flinewidth (0. pl_pencolorname ("black").phi. phi=atof(argv[5]).h> #include<math. } pl_fspace (-4. Ay=atof(argv[3]).cmd[3]. char *argv[]) { char Axl[14]="A\\sbx\\eb=".#include<stdio. Wx=atof(argv[2]). /* f[n] is a sine function for cuts */ float dt. int n. char phil[6]="\\*f=". stdout.h> #include<strings.h> #define N 1000 #define PI 3.1415927 float X[N].0.0. pl_selectpl (pl_handle). main (int argc.005). 4./lissojou Amp_x omega_x Amp_y omega_y phase display\n"). if(argc !=7){ printf(". Wy=atof(argv[4]).t.m.} Ax=atof(argv[1]).Ay. dt = 2. 4.l. char Wyl[14]="\\*w\\sby\\eb=".*cmdptr.Wx. int pl_handle. -4. if (pl_openpl () < 0) /* open Plotter */ { fprintf (stderr.Y[N]. stdin. char Ayl[14]="A\\sby\\eb=".0.k. 50 .Wy. pl_label(phil). labels=gcvt(Wx.n<999.15.buffer). -3.4.8.3.Y[0]). pl_label(Axl).0.labels). 51 .8).5.5).0. labels=gcvt(Ay.25). pl_fmove(0. pl_fmove(-3. pl_label(Wyl).4.3.labels). pl_fline(X[999]. pl_ffontsize(0. pl_closepl (). strcat(Ayl. pl_label("y(t)").buffer).4. pl_label(Ayl). strcat(Wxl.buffer). X[n]=Ax*sin(Wx*t).buffer).labels).labels).8).5.5).5).3.0).n++) pl_fline(X[n+1].8.labels).-0. -3.0. pl_fmove(-2. pl_fmove(3.buffer). -3. -3. for(n=0.0.4.X[0].X[n].4. pl_fline(-3. labels=gcvt(Ax.8. pl_selectpl (0). /* labels */ pl_ffontname("HersheySerif"). pl_fmove(-3. labels=gcvt(Wy.0.0. Y[999].n++){ t=(float)n*dt. strcat(phil.8).25). strcat(Wyl.7).4.Y[n]). pl_label("y’(t)"). strcat(Axl. pl_label(Wxl).0. -3. labels=gcvt(phi.0. Y[n+1]. } for(n=0. pl_fmove(0.pl_fline(0. pl_fmove(-2. -3.n<1000. Y[n]=Ay*sin(Wy*t-phi).15. pl_deletepl (pl_handle). } 52 . model 6218A. 53 . used as an ammeter.3 volt AC supply will provide current to a filament that heats the cathode of the diffraction tube. Power supply B. 4. the actual voltage value determines the electron momentum ac1 2 cording to eV = mc2 (γ − 1) ≈ 2m p The graphite target is powdered graphite bonded to a nickel mesh. a Teltron limited kV power supply.3 volt AC supply.1 Apparatus 1. Such a measure controls the number of electrons that will form the beam. The crystal lattice will form a diffraction grating from which electronic waves will scatter. connected between the cathod can and anode. This is done in such a way that electrons with insufficient energy to escape this voltage difference are returned to the cathode. Powder is used so that many different scattering planes are presented to the beam.6 Electron Diffraction and Bragg Scattering The purpose is to not only measure typical inter-atomic spacings between atoms in a typical crystalline solid (graphite) but to study as well the wave properties of electrons. These electrons will be controlled the 60 volt DC power supply. A Keithly 175 Auto-ranging multi-meter. too many in the beam will burn a hole in the graphite target. connected to the cathode. Power supply A. resulting in a circular ring of scattered electrons striking a fluorescent screen in the tube. The electrons will be accelerated by the 0 − 5000 volt DC power supply. The diffraction tube (Tel-Atomic) The 6. This consists of both a 0 − 5000 volt DC and a 6. 3. it will eject electrons. The total current must be kept below 100µA to prevent damage to the target. model 813. Once the cathode temperature is high enough. 2. a Hewlett-Packard 0 − 60 volt DC power supply. The layer is extremely thin. 6. 3. with inter-atomic spacings d1 .40 ˚ A. parallel planes of benzene rings. with an interplanar separation of about 3.3 VAC Cathode can Cathode r θ x D L Graphite target Electron beam 0-60 V DC Anode I 0-5000 V DC 6.54 ˚ A and a ˚ double bond of 1.5 A ˚ that perfectly superimpose. separated by planes There are planes separated by about 7A halfway in between.33 A.6.5 ˚ A. which is intermediate between a single carbon-carbon bond of 1.2 Ammeter Background Graphite is a carbon allotrope consisting of very large. This distance is about 1. These half-way planes have Carbon atoms in locations that superimpose 54 . on the centers of the hexagons in the planes above and below. by Ashcroft and Mermin.55˚ A with intercarbon spacings in each plane of 1. The optical path difference between the two beams must be an integer number of wavelengths in order that the beams constructively interfere when they reach a detector. and the inter-planar spacing is d = d1 sin α 55 d = h cos α . It is these two reflected beams that interfere. some transmitted to reflect from the next plane. The angle of incidence of the beam with respect to the plane normals must be just right in order that beams reflected from consecutive planes interfere constructively by time they reach the detector (glass shell of the diffraction tube). and an interleaving (red) plane in the figure. See Solid State Physics. 304 for details. a lower (blue). The stereogram below may help you see this. each separated by 3. x = h sin α. From the geometry alone we see that s = 2x sin α. this path condition is 2h − s = m λ according to the figure. the spacing is too big for that.4˚ A to scale. some of the beam is reflected. p. The spacings between atoms do not serve aperatures through which the electrons diffract. This condition is called the Bragg condition From the figure we can see that when the incoming beam arrives at the first plane of atoms. There is an upper (green) plane. What we see instead is reflection of the electron beam from consecutive planes of atoms. and θ is the scattering angle.incoming e to screen α α s d α α α d1 h x θ=π−2α from which we obtain mλ = 2h − 2x sin α = 2h(1 − sin2 α) = 2d cos α = 2d1 sin α cos α or mλ = d1 sin 2α = d1 sin θ in which d1 is the spacing between atoms. The voltage used to accelerate the electrons gives the kinetic energies that are nowhere near relativistic. and so we can use p = mv. We will look for the m = 1 bright interference band. the angle between incoming and outgoing beams. the primary. 1 2 1 p K = mv 2 = 2 2m and the de’Broglie condition λ= h p together with conservation of energy 1 eV = mv 2 2 to obtain the electron wavelength in terms of the accelerating voltage 56 . We process the data by first put this data into a file as follows 57 . the radius of curvature of the bulb. If we plot √1V versus sin θ.59 4.9 3.0 50.2 1.0 48.1 h 1.00 6.7 34.0 42.05 3.3 We have recorded the diameters of two rings.6cm. the inter-atomic spacing.93 3.0cm. and L = 14.22 3. the slope will give us d1 .77 5.8 2.08 5. and angular displacement of the diffraction rings.1 2. 6. For this apparatus we find the physical parameters are r = 6. the length of the bulb.3 Data and analysis Dinner cm 2.44 3. inner and outer. Examine the picture of the apparatus.5 2. We can measure the scattering angle θ by measuring the radius of the ring that the scattered electrons leave on the fluorescent tube.9 r 2 − ( D2 )2 ) We see that we must produce a plot of sin θ versus √1V and run a linear regression on the data to obtain the slope of the line of best fit in order to get the inter-atomic spacing.79 2.51 V kV 3. Notice tan θ = and x= s D 2 L−r+x r2 − ( D 2 ) 2 or D θ = tan−1 ( L−r+ q2 Douter cm 4.6 52.68 2.4 3.31 5.59 6.69 I µA 40.2268 × 10−9 mV olt 2 √ λ= √ = 2meV V we find that after using the Bragg condition s 1 2me √ = d1 sin θ h2 V is the relation between accelerating voltage and the scattering angle for the electron beam.7 39. 0943793 0.56+($1)*($1)/4. 1/sqrt(1000.0943793 0.5 2. awk ’{ print sin(atan2(($1)/2.120078 0.112709 0.1 2.106975 0.79 2.0179605 0.0213201 0.0160128 0.0229416 The graph of this is shown below.0179605 0.77 5.0981397 0.02 0." ".8 2.102907 0.68 2.2 1.69 4.93 3. 7.128375 0.00 6.0*$3)}’ data which results in the data for the inner ring 0. and linear regression 58 .0.120078 0.0171499 0.4 3.59 6.0213201 0.0188982 0.31 5.08 5.106975 0.02 0.0229416 which we can LaTeX format as a table Inner ¯ sin(θ) 0.05 3.0160128 0.0171499 0.128375 ring √1 V 0.0188982 0.112709 0.44 3.0981397 0.4+sqrt(43.2.59 4.102907 0.9 called “data” and run it through several Awk scripts to compute the average θ.51 3.0))).22 3.9 3. 8 %.0.10 0.14367 × 10 with an accepted inter-carbon in-plane separation of 1.018 0.198905 d1 = = 0.14 results in a line of best fit with √1V = −0.11 0.21 1.022 1/√V 0.020 0. χ2 = 0.44 Angstroms 8 8.019 0.017 0.12 sin(θ) 0. it is pssible that this ring represents scattering off of the large parallel planes of benzene rings.Inner ring 0.13 0.21 ˚ A separation illustrated below 1.4 for which our results are in excellent agreement of 0. for which our value is in error by 31 %.40 Angstroms.021 0.09 0.198905 sin θ . From this we find that 0.016 0. For the second ring we run 59 .002513 + 0.0244 × 10−8 m = 2. It is more likely that we are measuring twice the 1.023 0.000007 which is a very good fit. 14 0.0188982 0.017 0.019 0. √1 = −0.56+($2)*($2)/4.200617 0. resulting in V d2 = 0.023 0.018 0.117366 = 0.022 1/√V 0.179941 0.163204 0.16 0. 7.215368 0.14367 × 108 which is evidently scattering off of a different set of planes.0160128 0.22 .44 Angstroms 8.188429 0.157521 0.002244 + 0.17287 0.02 0.20 0.0144 × 10−8 m = 1.016 0.020 0.021 0.0. Outer ring 0.4 ˚ A.18 sin(θ) 60 0.0229416 and run the linear regression for the line of best fit.117366 sin θ with χ2 = 0.awk ’{ print sin(atan2(($2)/2. 1/sqrt(1000.0*$3)}’ data resulting in the data Outer ring ¯ √1 sin(θ) V 0.0213201 0.0))).4+sqrt(43.0179605 0. and is a good value for the in-plane inter-carbon spacing of 1.000008." ".0171499 0. There are two cells in these figures. m. 6= 90 Hexagonal 1 P a = b 6= c. which is the rectangular cubic structure. α = γ = β < 120. I=bcc. n. α = 90 = γ = β Tetragonal 2 P. I a = b 6= c. the spatial structure is invariant under translations by vectors T = na + mb + pc. α = 90 = γ 6= β Orthorhombic 4 P.4 Crystal structure determination by scattering In a crystal. the translation vectors T are illustrated below. α = 90 = γ = β Cubic 3 P=sc. α = 90 = γ = β Trigonal 1 R. I a = b = c. γ = 120 b α c γ β a For the important example of a body-centered cubic structure. C.6. C a 6= b 6= c. p integers forming a periodic structure falling into 14 types called Bravais lattices. the conventional cell. I. The notation for this is usually an ordered triple with no parenthesis m n p or commas. The atoms in the crystal form a periodic structure that superimoses upon itself if it is moved by any of these translation vectors. F a 6= b 6= c. System Triclinic 14 Bravais lattices Number of types Symbol Characteristics 1 P a 6= b 6= c. 61 . α = 90 = β. α 6= β 6= γ Monoclinic 2 P. F=fcc a = b = c. The smallest volume contained by this structure is the primitive cell. A stereogram of the primitive cell for a face-centered cubic lattice (in red) is illustrated below.04 ˚ A and an atomic nearest neighbor distance of 2. drawing a connecting line to each nearby lattice site. Solid state physicists propagate some confusion by reporting atomic positions for cubic lattices using the conventional cell coordinates. If you look at these pairs correctly.b c b c a a and the primitive cell which is the parallelopiped defined by the three lattice vectors a. 62 . which has volume |(a × b) · c|. and bisecting each of these lines with a plane. Aluminum is a good example of a face-centered-cubic structure with the length of a side of its conventional cell being 4. you will be able to see the three dimensional structure of the crystal. c and c. The Wigner-Seitz primitive cell is gotten by picking a lattice site. A stereogram of the primitive cell for a body-centered cubic lattice (in red) is illustrated below.86 ˚ A. Be careful not to break your brain. Consider waves of k-vector k = (0. for our example this 3 2 3 is (232). Atoms can be placed at various points within them. arriving at the observation point is Ψ(n. 12 12 0. In applications of these rules. when the atoms are merely separated by these vectors. 0 21 0. namely 1 1 1 .m. p) = eik·T relative to the wave arriving at the atom at the origin.p = (T − Rn) · (T − Rn) ≈ R2 − 2Rn · T. and a distance R away. The phase of the wave scattering from the atom at T. 0 12 12 and Chlorines at 21 21 12 . 12 00 in conventional cell (Cartesian) coordinates. The Miller indices of planes of atoms appear in the scattering formulas for these waves. These atoms and there positions in the conventional and primitive cells are illustrated in the stereogram below. 2b.p ≈ R − n · T . p) = eik·T+ikrn. Sodiums being at 000. The waves will arrive with a phase of ψ(n.63 ˚ A. 12 0 12 . which is fcccubic. Consider for example families of planes of atoms parallel to a plane defined by three atoms that lie on it.m. and constructing the smallest set of integers in the same ratio. This triplet is called the set of Miller indices. m.p = Rn Then 2 rn.p is the vector between the atom at T and the distant point. Consider spherical waves arriving at a distant observation point in direction n from the atom at the origin. .Once primitive cells have been established. 63 rn. so that T + rn. with the convention that the center of a primitive cell is at 21 12 12 . 0. The length of the side of the conventional cell in the NaCl structure is 5.p in which rn. 00 21 .m. each primitive cell contains four Sodium and four Chlorine atoms. m. Various types of waves incident on crystals interact with the atoms at lattice points. For example. for example 3a. in a NaCl crystal. a plane that intersects a crystal axis at infinity has the corresponding index equal to zero.m. A common error that people make is to think that the atoms of the lattice are located on the vectors T. 2π ) incident on an array of atoms at positions T = λ ma + nb + pc. The Miller indices of the plane are gotten by taking the reciprocals of these integers. and with planes upon which families of atoms lie. and 3c.m. 5 Simulation of scattering A typical crystal diffraction pattern for waves scattering off of a single crystal has bright spots in directions in which these three conditions are met. If we use the sum M X p=0 e ipA 1 − ei(M +1)A . or diffraction gratings. m. this amplitude becomes very sharply peaked around its central maximum. very reminiscent of the amplitudes for multiple slit interference. We can easily simulate this. (k − kn) · a = 2πma . 6. mc are integers. (k − kn) · c = 2πmc in which ma . As M becomes very large. p) = ei(k−kn)·T+ikR We now add up all of the scattered waves coming from all of the atoms in the lattice from which the waves scatter. (k − kn) · b = 2πmb .p =( X eim(k−kn)·a+ikR )( X ein(k−kn)·b+ikR )( = eiN kR ( X eip(k−kn)·c+ikR ) p n m X eim(k−kn)·a )( X ein(k−kn)·b )( eip(k−kn)·c ) p n m X Each sum appearing in this formula is in fact a geometrical series. being just a product of three diifraction grating amplitudes. in other words as the size of the crystal that we are scattering off of increases. For example. The maximum occurs for n directions for which all three factors in the product are one.and our single arriving wave has phase Ψ(n. X i(k−kn)·(ma+nb+pc)+ikR Ψ= e m. and in fact are equivalent to the Bragg condition. the (photographic negative) of the diffraction pattern for scattering from a 9×9×9 hexagonal.n. These are the Laue conditions. mb . close packed crystal looks like the following 64 . = 1 − eiA | M X e ipA p=0 sin( M2+1 A) |= sin( A2 ) we discover a wave amplitude of |Ψ| = sin( M2+1 (k − kn) · a) sin( M2+1 (k − kn) · b) sin( M2+1 (k − kn) · c) sin( (k−kn)·a ) 2 sin( (k−kn)·b ) 2 sin( (k−kn)·c ) 2 in which M is some large integer. In polycrystalline scattering, like off of a powder consisting of many single crystals with arbitrary orientations, the diffraction pattern is similar to that above, but is gotten from the spot pattern by rotation to form concentric rings. Below you will find a simple C program for simulating this experiment and testing the Laue conditions. You may make changes in it, to output the angles θ and φ of the bright spots, so that you can verify the conditions by hand. #include<stdio.h> #include<stdlib.h> #include<math.h> #include<plot.h> #define PI 3.1415926 __complex__ double phase,eye; __complex__ double cexp(__complex__ double num); double modulus(__complex__ double num); int m,n,p,M,N,pl_handle,num; double a[3],b[3],c[3]; double r[3],Q,R,S; double k[3],kp[3],spot[1000][3]; double phi,theta,x,y,dx,dy,amp,R,lambda; main(){ /* lattice vector components */ a[0]=2.0,a[1]=0.0,a[2]=0.0; b[1]=-sqrt(3.0),b[0]=1.0,b[2]=0.0; c[2]=2.0,c[0]=0.0,c[1]=0.0; eye=0.0+1.0i; dx=0.05; dy=0.05; /* x-ray wavelength */ lambda=0.05; k[2]=2.0*PI/lambda;k[0]=0.0;k[1]=0.0; 65 pl_handle = pl_newpl ("ps", stdin, stdout, stderr); pl_selectpl (pl_handle); if (pl_openpl () < 0) /* open Plotter */ { fprintf (stderr, "Couldn’t open Plotter\n"); return 1; } pl_fspace (-4.0, -4.0,4.0, 4.0); pl_filltype(1); pl_fillcolorname("white"); /* pl_fbox(-4.0, -4.0,4.0, 4.0); */ pl_flinewidth (0.005); for(M=-40;M<40;M++){ for(N=-40;N<40;N++){ /* using over 10,000 points on viewing screen */ R=sqrt(((double)N*dx)*((double)N*dx)+((double)M*dy)*((double)M*dy)); theta=atan2(R,30.0); phi=atan2((double)M*dy,(double)N*dx); /* this is k’- the elastically scattered wavevectors */ kp[0]=(2.0*PI/lambda)*cos(phi)*sin(theta); kp[1]=(2.0*PI/lambda)*sin(phi)*sin(theta); kp[2]=(2.0*PI/lambda)*cos(theta); phase=0.0+0.0i; /* lattice will be 9x9x9, pretty big */ for(m=-4;m<=4;m++){ for(n=-4;n<=4;n++){ for(p=-4;p<=4;p++){ /* generate a lattice scattering site */ r[0]=(double)m*a[0]+(double)n*b[0]+(double)p*c[0]; r[1]=(double)m*a[1]+(double)n*b[1]+(double)p*c[1]; r[2]=(double)m*a[2]+(double)n*b[2]+(double)p*c[2]; phase=phase+cexp(eye*(r[0]*(k[0]-kp[0])+r[1]*(k[1]-kp[1])+\ r[2]*(k[2]-kp[2]))); }}} amp=modulus(phase); /*printf("%f\t%f\t%f\n",(double)N*dx, (double)M*dy, amp);*/ num=(int)(65535.0*(amp/729.0)); if(num>4000){ pl_color(num,num,num); /* output a spot (hopefully) corresp. to reciprocal wavevector */ pl_fbox((double)N*dx-dx/2.0,(double)M*dy-dy/2.0,(double)N*dx+dx/2.0,(double)M*dy+dy/2.0); /* check for laue conditions */ Q=(a[0]*(k[0]-kp[0])+a[1]*(k[1]-kp[1])+a[2]*(k[2]-kp[2]))/(2.0*PI); R=(b[0]*(k[0]-kp[0])+b[1]*(k[1]-kp[1])+b[2]*(k[2]-kp[2]))/(2.0*PI); S=(c[0]*(k[0]-kp[0])+c[1]*(k[1]-kp[1])+c[2]*(k[2]-kp[2]))/(2.0*PI); 66 /* output Q=m_a, R=m_b, S=m_c for brightest spots */ printf("%f & %f & %f & %f\\\\\n",100.0*(float)num/(float)65535.0,Q,R,S); } }} pl_closepl (); pl_selectpl (0); pl_deletepl (pl_handle); } __complex__ double cexp(__complex__ double num){ __complex__ double tmp; __real__ tmp=exp(__real__ num)*cos(__imag__ num); __imag__ tmp=exp(__real__ num)*sin(__imag__ num); return(tmp); } double modulus(__complex__ double num){ double tmpx,tmpy; tmpx=__real__ num; tmpy=__imag__ num; return(sqrt(tmpx*tmpx+tmpy*tmpy)); } The centers of the very brightest spots are organized in the table below, and the “integers” for the Laue conditions are displayed. You can see that even for a mere 729 atoms, the scattering maxima obey the Laue conditions quite well. 67 Laue bright spots % brightness ma mb 43.862058 0.998751 -0.999860 43.837644 -0.998751 -1.998611 57.804227 -0.000000 -1.038763 84.116884 -0.000000 -0.981102 82.714580 0.999575 -0.019607 83.109789 -0.999575 -1.019182 59.916075 0.999599 0.038104 59.301137 -0.999599 -0.961494 43.884947 1.997505 0.998752 100.000000 0.000000 0.000000 43.884947 -1.997505 -0.998752 59.301137 0.999599 0.961494 59.916075 -0.999599 -0.038104 83.109789 0.999575 1.019182 82.714580 -0.999575 0.019607 84.116884 -0.000000 0.981102 57.804227 -0.000000 1.038763 43.837644 0.998751 1.998611 43.862058 -0.998751 0.999860 mc 0.049962 0.049962 0.017988 0.016046 0.016989 0.016989 0.016046 0.016046 0.049906 0.000000 0.049906 0.016046 0.016046 0.016989 0.016989 0.016046 0.017988 0.049962 0.049962 Read through the program and determine how far the crystal sample is from the screen on which these spots are cast, and find the wavelength. Use the data to actually determine the lattice spacings. This might give you some insights into your experiment in electron diffraction. 68 3. not a lot) can buy.5 cm focusing lenses. catalog number 2120. we measure Planck’s constant and verify the particulate nature of light. Filters built into the apparatus restrict the frequencies of light entering the device to a very narrow bandwidth. Welch Scientific model 36 Universal light source. Sargent-Welch Scientific. 4. 7. 10 cm and 2. When the current is zero. Keithley 610c solid state electro-meter. Keithley 485 auto-ranging picoammeter. It provides an excellent contrast to the electron diffraction experiment that illustrates the wave nature of matter. to be sure that we know precisely which wavelengths cause photo-emission.5 volt size D batteries in series. 3 volt power supply. A voltmeter is used to exactly measure this anode/cathode voltage. 2. 2 1. no electrons have sufficient kinetic energy to reach the cathode. The lense is used to eliminate other light sources andto direct the beam from the universal light source. They are used to precisely control the anode/cathode voltage. A light source.1 Apparatus 1. 5. We really only want light of one frequency entering the device. It provides the photons that cause photo-emission of electrons from the target metal (anode). Voltmeter. The source itself is as close to white light as money (a little. The role of the ammeter is to measure the current of the photo-electrons. the interpretation of which Einstein received the Nobel prize. There are a pair of variable resistors R1 and R2 built into the Planck apparatus. The 3 volt power source provides a fixed anode/cathode voltage difference and will be used to measure the maximum energy of the ejected photo-electrons. 69 .7 The Photoelectric Effect In this experiment. Planck constant apparatus. 6. Ammeter. light incident on the anode can be absorbed by electrons . In terms of light wavelength this is hc λ The kinetic energy K can be measured by applying sufficient voltage to stop the electron K +φ= K = eVstop and we find that Vstop = φ h ν− e e 70 . and if they absorb enough energy.White light Lense Changeable filter Electric field Cathode Anode electron beam I Ammeter V Voltmeter R1 R2 3 V DC 7. The kinetic energy of the electron plus the energy required to liberate them from the anode must equal the energy of the photon. and naturally varies from metal to metal.2 Background In the Photoelectric effect. the electrons can be liberated from their potential wells and escape the metal. here assumed to be a discrete particle with energy related to its frequency K + φ = hν φ is called the work function. 0043 -.0042 -. 577 nm respectively) and determine the stopping voltage by looking for the place where the data curves start to deviate from the straight line portion.55 -.10 -.0061 -.0055 -.0043 We now graph this data ( voltage versus current.950 -.2375 .0034 -.0130 1.0018 .0036 -.35 -. 7.0404 .0041 -.0329 .0036 -.550 -.0373 .350 .0090 1.0001 .050 .45 -.0059 -.1768 .0134 1.0127 .3 Procedure and data We will apply various voltages to the anode/cathode and monitor the current for several light wavelengths.250 .0928 .3227 .75 -.0136 1.0066 .850 -. since no current passes due to flowing electrons once this point is reached.0016 1. 435.05 -.650 -.25 -.0143 2.0140 1.8 nm filter V I I I volt µA µA µA 0.0052 .0138 1.0041 -.65 -.8 nm.6568 0.0027 -.0060 -.50 -. 71 .00 -.750 -.450 .0119 1.0057 -.0062 -.0145 2.0042 -.5461 .0898 .6206 1. 546 nm.0040 -.75 -.0098 .0038 -.0058 -.3475 .3882 1.2070 .2009 .0058 -.15 -.0043 -.0143 2.8615 .0039 -. 577 nm filter 546 nm filter 435.stopping voltage versus light frequency is a straight line whose slope gives the value of Planck’s constant and intercept determines the work function of the metal.0045 .0971 .0056 -.150 .0037 -.0141 2. 5 2.25 0.2 0.8667 ± 0.00 −0.0 2.0 2.2.0 0.0866 volts and for the 435.5 2.8 nm (ν = 6.5 2.35 0.0 0.05 0.4 1.5 0.0 1.7 0.0 2.6 1. to get the slope of the best fit line 72 .0866 volts This we now graph.0 0.1 0.0 By doing a spline fit we find that for 577 nm light ( ν = 5.3 0.5 1.5 0.0 0.33 ± 0.0 −0.5 0.20 0.0 0.8791 × 1014 Hz) data Vstop = 1.5 3.5 0.4907 × 1014 Hz) data we obtain Vstop = .15 0.10 0. stopping voltage versus frequency or else do a linear regression on the data.5 1.30 0.1 0.05 0.5 1.0 1.0 1.7167 ± 0.5 0.1957 × 1014 Hz) Vstop = 0.0 0.5 −0.0866 volts for the 546 nm ( ν = 5. 356822 ± 0.032069 volt h = 14 10 Hz e from which we get h = 5. the multiply degenerate energy levels of the free atoms blend into energy bands. Hopefully you will be able to obtain five or even six data points for the Planck line. 8.5 7.0 5.4 4. in the solid.5 6.17163 ± 1.0 0.1 Theory When free atoms are brought together to form a periodic-lattice solid. and there is usually a gap between the valence and conduction bands. In addition you will need a 2 − 12 V AC variac and two voltmeters (DMMs).2 1. The presence of electrons or other charge carriers in the conduction band determine electrical conductivity of the solid. and bonding electrons often compose a seperate higher energy band. the conduction band. the valence band.5 5. close to the accepted value of 6.0913 × 10−34 J · sec which is a rather good result .0 6.1. The valence electrons of the seperated atoms tend to form into one band.8 0.0 The linear regression gives us a slope of 0. 73 .4 1. containing a selection of six LED (light emitting diodes) in series with a 100 Ω resistor. near continuous collections of electron levels.6262 × 10−34 J · sec 8 Measuring Planck’s constant (solid state version) In this version of the experiment you will use the Klinger LED apparatus.6 0. Holes and electrons then “annihilate” one another. on the n-side.Conduction band Eg Valence band Seperate atoms Solid state We know that an n-type semiconductor has electrons in its conduction band. p−type n−type E Conduction Valence Distance The gray areas represent filled electronic band states. Before an n-type and p-type are joined to form a pnjunction. 74 . and on the p-side the hole concentration exceeds that of the n-side. and holes diffuse from p-type to ntype. this forms a thin layer around the junction in which there will be excesses of the appropriate donor or acceptor ions. Once the two materials are physically joined. there are electron and hole concentration gradients at the junction. there is a higher concentration of electrons than on the p-side. and so electrons will diffuse from the n-type into the p-type. and a p-type has some holes in the valence band. the electron energy diagrams for the two look like the following. Concentration gradients drive diffusion. allowing an equilibrium the result. on the p-side an imbalance of electron donor ions. a constant Vn in the n-type.E p−type n−type Conduction − + Valence Distance This thin layer is called the space-charge region. This is called the built-in field. The field exerts forces on electrons and holes that opposes the diffusion. we see that a solution to the Poisson equation is a constant potential Vp in the p-type material. Examine the electrostatic potential energy of electrons on both sides of the junction. and the ramping potential due to the built-in field connects the two in the space-charge region. and this creates an electric field and potential in the space-charge region. since and outside of the space-charge region the two materials are electrically Ebuilt−in = − dV dx neutral. here illustrated with a Fermi level indicated by a dotted line. and on the n-side of it there will be an imbalance of acceptor ions. 75 . E V(x) p−type n−type Conduction band Vn EF Ebuiltin Valence band Vp Multiplication of this potential by the electron charge (negative) results in the electronic energy structure illustrated above for a pn-junction at equilibrium. where they combine with holes. and electrons can migrate into the lower levels of the conduction band on the p-side from the n-side. and a p-type solid with holes acting as charge carriers. the energies of the n-type bands will be raised to a point such that the barrier presented to the flow of electrons from the n-side is dramatically reduced. 76 .Consider the energy level scheme of a pn-junction forming the boundary between an n-type solid with a partially filled conduction band (filled electron states) up to some Fermi level (indicated on the right). there are no empty hole states in the conduction band on the p-side. On the p-side. empty electron states (holes) in the valence band are not shaded. and the filled hole states fall short of the valence-to-conduction band gap on the p-side. The gap is drawn on the p-side for convenience. E p−type n−type Filled e− states (n−conduction band) EF Empty states (holes) not shaded (p−valence band) This results in the creation of a photon whose energy is the band-gap energy(between valence and conduction bands). If the junction is forward-biased. the emitted photon will be visible. This process may be a bit easier to understand. combine with electrons and emit the same frequency light. In addition holes can migrate from the p-side to the n-side.5 eV . one can see that we have |e| Vmin = h ¯ ω = hν. the hole migrating to the n-type side is a vacancy in the valence band on the n-side. 77 |e| Vmin ≤ Eg . when the applied forward bias is at the threshold for a charge carrier transfer from one side to another.E hν=Egap n−type p−type |e| Vmin If the band-gap is at least Eg ≥ 1. which “annihilates” an electron by providing a vacancy that an electron in the conduction band can fill (to oversimplify the matter somewhat). E p−type n−type hν=Eg |e| Vmin In any event. which allows us to measure Planck’s constant and verify the Planck Hypothesis by measuring the threshold voltage for forward-biased conduction and plotting it versus LED dominant radiated frequency. 78 . and use the method of bisection program to find the roots or zeros of this function. Both programs can be obtained from the course website. The current will be very small. but the distribution is fairly sharp at a precise value. In the real world the diode emits a variety of frequencies. 8. for each diode a plot of current versus voltage across the diode needs to be produced. and use the computer program for fitting data to polynomial curves to fit that portion of your data for which I > 0 to a high order polynomial such as I = a4 V 4 + a3 V 3 + · · · a1 V + a0 . Example Suppose that I took the (ficticious) current versus voltage data below. This will give you Vmin in each case. at which point it will climb rapidly. 100 Ω mA V DMM Do this for each diode. The wavelengths of the dominant photons are printed right on the apparatus. but not zero (due to tunneling) until the voltage drop reaches the required Vmin for forward conduction.2 Procedure It should be fairly clear what needs to be done. 3 Amp 2 Amp 4 V + 0. obtaining I(V ) = −0. If there is . The simple C + + program below accomplishes this.7 Amp + 0. that with current equal to zero (squares) and that with current greater than zero (circles).5 V 2 V olt V olt4 as the best fitting quartic.0 V .2. If there was not. 2 . the fifth and sixth order fits being worse by a Chi-squared criterion. then there is a root between z and x2 and so we reset x1 = z and iterate. reset x2 = z and iterate.1 Finding roots by bisection To apply the bisection method. bisect the segment and let z = x1 +x 2 is a root between x1 and z.0 I have divided the data into two sets.0 0. Now test if there If there is a root between x1 and x2 . 8. 79 .5 1.7 6 I(mA) 5 4 3 2 1 0 0. If there is a root between x1 and x2 . I now run a least-squares polynomial fit on the circle-data. then f (x1 )f (x2 ) < 0 otherwise f (x1 )f (x2 ) > 0 The first step in the algorithm is to test which case prevails. I obtain Vmin by the method of bisection or Newton’s method.0 V 1. pick two points. x1 and x2 . We continue to do this until we have narrowed down the interval in which there is a root to whatever accuracy we desire. beginning with a guess near 1.5 2. 8.n++){ if (f(x0)*f((x0+x1)/2.} Once you have the least squares polynomial fit. other than stability. xn+1 should be closer.0.0. cout<<" Input x0 and x1. you can create a subroutine that computes the polynomial. and point the function pointer to your routine.0) x1=(x0+x1)/2.n<20.2 Newton’s method The Method of Newton is another iterative scheme for finding or refining roots that can be potentially unstable. and so should be used with caution.0)<0. If f (xn+1 ) ≈ f (xn ) + (xn+1 − xn )f 0 (xn ) ≈ 0 then xn+1 ≈ xn − f (xn ) f 0 (xn ) so if xn is close to a root.2.0*x+1. is that you need the derivative function as well as the function itself in order to evaluate this expression.#include<iostream. for(int n=0.0). If you have both functions.h> float f(float x).h> #include<math. cout<<"the root is = "<<root<<"\n". cin>>x0>>x1. this can be used to polish or refine an estimate of a root gotten by bisection. 80 . } float root=(x0+x1)/2.x1. two points around a root"<<"\n". The disadvantage.0. } float f(float x){ return(x*x-3. else x0=(x0+x1)/2. /* define your function in this routine */ main() { float x0. y. return.One distinct advantage of the Newton method over bisction is that since it is non-geometrical. it applies equally well to real or complex roots. otherwise polynomials with real coefficients are always real-valued. 81 \ .0e-15). do{ y=x0.y. For example /* Newton-Rafeson computation of roots of cubic */ /* for complex root.x0). __complex__ double x){ __complex__ double del.__real__ x0.0). } __complex__ double f(__complex__ double x){ return(x*x*x+1. \ __complex__ double x). __complex__ double f(__complex__ double x).9i.x1.0+0. /* derivative routine accepts pointer to function */ double modulus(__complex__ double z). a complex initial guess is used.h> int n.}while(modulus(y-x0)>=1.__imag__ x0). x0=x1. /* pass derivative a pointer to f */ printf("%f+%fi\n". } __complex__ double df(__complex__ double (*pf)(__complex__ double x). One needs to make sure that if a complex root is sought. make sure x0 (first guess) */ /* is a complex number */ #include<stdio. main(){ x0=1. __complex__ double x0. /* test for precision */ x1=x0-f(x0)/df(f. __complex__ double df(__complex__ double (*pf)(__complex__ double x). im. 82 . return((re*re+im*im)).0*del). and you can devise other methods as well. along with a least squares fit to determine Planck’s constant from the slope. return(y). You will also report a table of Vmin versus diode wavelength and frequency or each diode. and an error estimate for it. graphs of all six sets of data. The lab computers are equipped with CYGWIN.\ (*pf)(x-2.0*del))/(12. re=__real__ z. You must compare your value of Planck’s constant with the accepted value.0*del). y=(8. and on CYGWIN or MINGW32 as well. im=__imag__ z. and the actual polynomial.del=0. and finally produce a graph of Vmin versus f = ν. } At any rate these programs give you at least two options for computing Vmin from I(Vmin ) = 0. Your lab report will contain the current versus diode voltage drop for all six diodes in tabular form.0*(*pf)(x+del)-8. fourth or higher order. and LaTeX for your convenience.1.0*(*pf)(x-del)-(*pf)(x+2. All of these programs compile on any Linux or FreeBSD computer. the set of points used for the polynomial fit must be clearly marked since I will perform the fit to check your work. } double modulus(__complex__ double z){ double re. Then an inelastic collision takes place. What we are seeing is 83 . and so we expect the collector ring current to increase with accelerating voltage. The electrons are first accelerated by a voltage Vc between the apparatus cathode and a grid.9eV for mercury. which measures the energy of electrons arriving at the anode. In this case they arrive at the anode with a final energy Ekin = e(Vc − Va ) − ∆E. the Mercury atom is excited. we see fluctuations in the current at precise intervals. electrons arrive at the collector ring in the apparatus at an increasing rate. As anode current. The Franck-Hertz Experiment How do electrons interact with ordinary atomic matter? In 1914 J. Hertz performed an experiment in which a beam of electrons is passed through a gas of atoms. Hg sample 25 I milliamps 20 15 10 5 0 0 5 10 15 V. and the ring current drops since electrons are not reaching it. Electrons should arrive at the anode with energy e(Vc − Va ). and also by a decelerating voltage Va between the anode and grid.9 Helium Ionization Potential. This indicates that there is a threshold energy for electron/atom interactions. and that the thresholds occur at energies regular intervals of 4. Some of the electrons will lose energy to the atoms (originally a gas of mercury vapor) . volts 20 25 As the accelerating voltage is increased. is monitored as a function of Vc .Franck and G. This does occur. until the incoming electrons have enough kinetic energy to excite the Mercury atom in the tube. unless of course they interact with the gas atoms. the incoming electron virtually stops dead. 5. we excite a gas of helium atoms with the electron beam. Franck-Hertz tube. This is very strong evidence for the discrete nature of bound electron energy states. Keithley 175 auto-ranging multi-meter. Ammeter. 9. 4. We perform a variation on the classic experiment here at Parkside. 3. Rheostat. 6. Single 1. 7.1 1. This particular apparatus may be phased out. This is reinforced by the fact that each 4.4 amp . Signal generator. 84 .9eV energy loss is accompanied by photon emissions from the mercury. 4.5 volt size D battery. 12 volt power supply. 12 volt automotive battery.5 V power supply. Wavetek model 182A function generator. 1. catalog number 8291013. A short description with data for a newer apparatus follows this section. Instead of mercury. 8. 20 V power supply. Voltmeter. 2. Heathkit IP-2718. Tektronics Franck-Hertz apparatus. Cenco 22 ohm. Keithley 178 digital multi-meter. since it is a bitch to set up. apparatus 1 MegaOhm resistor. Cornell-Dublier electronics decade resistor model RDC.the projectile electron giving up energy to promote bound electrons in the mercury atom to higher atomic energy levels. The old version of the experiment. of light with a wavelength corresponding to this energy difference. 9. Farm’n Fleet model UL-300CA. from 20 to 26 volts. and the voltage across the resistor.5 volt battery is used to raise the collection ring voltage slightly above the anode voltage.These are assembled as shown below. through which runs the current of electrons that arrive at the collector ring. Increases in the electrons making it to the ring increase the current. 20 volt power supply is used raise the saw-tooth voltage from the signal generator to such a point that the peak to peak value of the total voltage input spans the full energy range for excitation of the helium. Oscilloscope monitors the voltage across the 1M Ω resistor.5 V DC 0-20 V power supply Oscilloscope We briefly describe the role each item plays in the experiment. which needs to be kept below 5 volts 85 . The scope also monitors times at which the voltage reaches a value that indicates absorptions by the helium gas. 1. Rather than performing the entire experiment with several cathode voltages. decreases in electron flow translate into drops in beam kinetic energy due to energy absorption by the gas.5 amps without being damaged. The ammeter monitors the filament current. we perform it with an AC sawtooth voltage that sweeps through the entire excitation range periodically. The resistor is in the circuit simply to aid in measuring the voltage or current from the collection ring. The voltmeter monitors voltage across the filament. Electrons that lose energy in collisions with helium atoms will be moving slowly enough to be collected. This is signaled by sudden drops of voltage across the resistor. which cannot exceed 1. Ammeter collecting ring Filament A anode Rheostat V Voltmeter + 12 V Power supply 6 V peak-to-peak signal generator 1M - + 1. The signal generator provides a predictable voltage that increases at a determined rate. A battery is more constant than an electronic power supply.05 ± 0.05eV ms V E3 = e(0. These will be accelerated and will collide with helium in the tube.5 ± 0.1.4 6.55eV.8 ± .87ms) = 1.1ms times 2 since the scale on the scope has 2 ms per division. E3 = 4.2 Data and analysis Once the apparatus is assembled and all voltages are in the required range.789 V )(1.1 ± 0.2 ± 0.9 ± . no ripples or fluctuations. we need to know the slope of the signal generator waveform ∆V = 6.48eV ms V )(3.025 V ms and we can compute the energies of the excitations E1 = e(0. 3.8 eV respectively. Varying the voltage on the anode creates the accelerating potential difference for the emitted electrons. 86 .4 3.4 6. 9.1. the commonly accepted values.789 ± 0.6eV. This produces a constant production of thermionic electrons.1 volts ∆t = 3. Another safety precaution.in order not to damage the filament.4 1.1ms) = 4.81eV ms and we can perform standard Gaussian error analysis on our data to arrive at E2 = e(0. We adjust the 20 V power voltage until three excitation peaks are visible on the oscilloscope.789 E1 = 1. This gives us a slope of 0.0 ± 0.4 1. we can take data.9 ± 0. which are collected by the collector ring.4 In order to transform these times into voltages. E2 = 3.9 ± 0.87ms) = 3.8 ± 0.81 ± 0. The Franck-Hertz tube contains a filament from which electrons are expelled by thermionic emission.8 ± 0. 12 volt power supply provides a constant DC voltage to the filament.9 ± 0. The rheostat limits the flow of current to the filament. 4.4 6.57eV which agree well with 1.4 3.0 ± 0. again to prevent burn out. and the time of the peaks are recorded in milliseconds E1 E2 E3 trial 1 trial 2 trial 3 ms ms ms 1.4 3.789 )(6. corresponding to Neon atomic absorptions of 18. 38. otherwise the collisions are elastic. 21. 39.5.02 Neon tube. Neon sample 50 40 I 30 20 10 0 0 10 20 30 V.93 eV . This energy loss equals the difference in two atomic energy levels of the Neon. Remember that it is the sudden drop in current at the peak that signals the prpjectile electron energy loss. and 41.0. and 41.22. 21.0. otherwise it functions pretty much the same as in the previous description.30. it may reach another critical value and reveal another atomic energy level. or the This version uses the TEL 2533. The tube containing the gas sample has a weakly conducting inner coating connected to the anode. volts 40 50 recording collector ring current (from electrons that give up all or most energy to excite atoms) versus accelerating voltage.08. 87 . a typical chart recorder output looks like this. 34. You can see this with a diffraction grating.3 Second version This version uses the TEL 2533. Recommended as well are the TEL 2800 LV and TEL 2801 HV DigiRamp power supplies.08. and a picommater. and the Neon will then spontaneously decay by emitting light at these energies.5 V .30.93 V . As the beam undergoes further acceleration by the field in the tube. 39. For the Neon tube. and a chart recorder. accelerated electrons undergo inelastic collisions with gas atoms once they reach sufficient energy to cause the atoms to undergo an atomic transition.9.22. These compare very favorably with the published peak voltages of 17.01 Helium Hertz critical potential tube. In this figure the current peaks at 18. 21. 5 eV corresponds to Neon ionization by the incoming electron. I would suggest determining which spectral line is being excited for each of your peak dropoffs. Neon has spectral lines at λ = 73.7 nm. The second peak at 21. if the atom were to shed this energy as light.2. 585. 73.8 nm. The first peak drops off rapidly at E = 18. 88 .30 eV .2. Correcting for the contact potential will improve the accuracy. which compares well with the known Neon line at λN e = 73. it would be at wavelength 1240 eV · nm hc = = E = 18.5.30 eV λ λ or λ = 67.Examine the first pair of peaks for this run.4. 640.5 nm. 671. 1 Background The “Old Quantum Theory” that supplanted classical mechanics as a description of the electron in an atom was based We now state the previously derived quantization rule as a H postulate Postulate W = p dq = nh where h is Planck’s constant. which we evaluate using the Cauchy theorem V (r) = 89 . Apply this now to the problem of finding the quantum energy levels of the hydrogen atom. by eliminating the derivatives p2φ p2r p2 + V (r) + θ2+ 2m 2mr 2mr 2 sin2 θ Impose the quantization conditions E=H= or pθ = `¯ h and similarly pφ = m¯ h and I pr dr = I s I pθ dθ = `h 2m(E − V (r) − (`¯ h)2 ) dr = nh 2mr 2 Now insert the form of the potential k r and note that the contour integral has two poles. the theory of which we have studied in Physics 202. The grating will be incorporated into a spectrometer. with lagrangean m L = (r˙ 2 + r 2 θ˙2 + r 2 sin2 θ φ˙ 2 ) − V (r) 2 The momenta are pr = mr˙ pφ = mr 2 sin2 θ φ˙ pθ = mr 2 θ˙ and the Hamiltonian is then. 10. you will perform the same experiment. and we will study the measure the Rydberg constant for hydrogen.10 The Rydberg Constant This constant determines the spacing between Hydrogenic Bohr energy levels. one for large r and one for small. In our modern lab experiments course. This simply requires a diffraction grating. and was originally determined spectroscopically. but only one term will have a residue I I q 0− s I √ m k k + · · ·)dr = −2πik 2m(E − )dr = 2mE (1 − r 2Er 2E r The final result is m = nh 2E This can be inverted to get E. For small r we neglect all but the last term under the radical to get s 2m(n¯ h)2 = 2πi −`2 h ¯ 2 = −`h 2mr 2 For large values of r we instead throw out the last term in the radical and expand using the binomial theorem. k 90 s m = (n + `)¯ h 2|E| .I I dz = 2πi z z k dz = 0 . These states differ in their orbital eccentricity. recall that E is in fact negative I pr dr = −`h − 2πik r −mk 2 E= 2 2¯ h (` + n)2 The interpretation of this result is very simple. From our classical mechanics experience we know that planetary orbits are elliptical with a semimajor axis a given by k a= 2|E| and semiminor axis b determined by both energy and angular momentum b=q L 2m|E| such that the orbital eccentricity is = s b2 1− 2 = a s 1− 2|E|L2 mk 2 For the Sommerfeld calculation these become q h ¯ ` = b 2m|E|. there is an energy level degeneracy due to the fact that several different states with different angular momenta can have the same energy.for any k value other than −1. orbits that would be degenerate (have the same energy) without relativity 91 .53 Angstroms is the Bohr radius. Some of these orbits are illustrated below b= N=1. l=1 Sommerfeld was able to go one step further. E= −mk 2 2¯ h2 N 2 This means that our orbits have quantized semimajor and semiminor axes. k h ¯2 2 aN = N = a0 N 2 = 2|E|N mk and ` aN = `N a0 N in which a0 = 0. all of the integrations are pretty much the same. l=2 N=2. l=3 N=3.Multiply these. by beginning with a relativistically correct expression for the energy. l=1 N=3. l=1 N=2. l=2 N=3. and the end result illustrates the fine structure splittings in the energies of certain orbits with the same N but different `. h ¯` q b 2m|E| or k s m h ¯ `a = = (n + `)¯ h 2|E| b n+` N a = = b ` ` in which the energy quantum number is N = n + `. We conclude that relativistically s `2 c2 h ¯2 k k2 1 (E 2 + 2E + 2 ) − m2 c4 − pr = c r r r2 and therefore its corresponding action quantization is I s 1 `2 c2 h ¯2 k k2 (E 2 + 2E + 2 ) − m2 c4 − dr = nh c r r r2 This is exactly what we had before. with a few changes.taken into account. Integrating exactly the same way as before we obtain s k2 Ek i 2 − `2 h ¯2 + i √ 2 = h¯ h c c E − m 2 c4 If we call k Zc¯ h = e2 4π0 h ¯c = α. the fine structure constant. we can rewrite this as √ EZα =n i Z 2 α 2 − `2 + i √ 2 E − m 2 c4 and we expand in powers of the fine structure constant. √ and so i√ Z 2 α2 − `2 Z 2 α2 + · · ·) ≈ i(` − 2` Z 2 α2 Z 2 α2 EZα ≈ (n + `) − =N− 2` 2` E 2 − m 2 c4 92 . k 2 p2θ 2 4 2 2 (E + ) − m c = c (pr + 2 ) r r but again I pθ dθ = `h and since pθ is conserved (constant) this gives us pθ = `¯ h again. The total energy according to relativity is E= q c2 p~2 + m2 c4 − k r which we write as k (E + )2 − m2 c4 = c2 p~2 r and we now insert or old expression for the momentum in polar coordinates from our first Sommerfeld computation. which is very small. with only the proviso that ` is replaced by `+ 21 to correct for electron spin. The basic electron energy is mc2 = 511 keV .23eV . 93 . and all of the rest is the quantized electronic energy EN. even when compared to computations performed with the relativistic version of the Schrodinger equation (the Dirac equation). and these fine structure corrections are four orders of magnitude smaller still. yet this expression for the fine structure spectrum is exactly correct.451 × 10−3 eV . or keV mc2 α4 = 511 ≈ 1.0072974 ≈ 4π0 h ¯c 137 and this provides us with a very valuable tool for getting an idea as to how large a correction to the energy each term represents.02723 keV = (137)2 27.or m2 c4 = E 2 (1 + Z 2 α2 ) 2 α2 )2 N 2 (1 − Z2`N This gives us the final relativistically correct energy quantization rule E=r ≈ mc2 q mc2 1+ Z 2 α2 2 α2 2 N 2 (1− Z2`N ) 1 1+ Z 2 α2 N2 + Z 4 α4 N 3` +··· 1 Z 2 α2 Z 4 α4 3 Z 2 α2 ≈ mc2 (1 − ( 2 + 3 ) + ( 2 )2 + · · · ) 2 N N ` 8 N 2 2 2 2 Z α Z α 1 3 ≈ mc2 (1 − (1 + ( − )) + · · · ) 2N 2 N ` 4N The first term we recognize as the rest energy of the electron.451 × 10−6 keV = 1. The Bohr keV orbitals represent states whose energies differ from this by α 2 mc2 = 511 ≈ 0. The fine structure constant is α= e2 1 = 0. a factor unknown at Sommerfeld’s time.` = − mc2 Z 2 α2 mc2 Z 4 α4 1 3 − ( − )+··· 2 3 2N 2N ` 4N This shows that the different ` elliptical orbits allowed for a given N value are not degenerate but are split by some very small energies. A most remarkable fact is that we (137)4 know that Bohr-Sommerfeld theory is very flawed. 38 eV Balmer Transitions (vis) Lyman Transitions (UV) N=1 -13. When this occurs.2 Balmer Atomic Spectra By hypothesis. There can be six wavelengths of light emitted when a gas of such atoms is thermally excited.E N=4 N=3 -0. light of frequency f and wavelength λ is emitted with hc = EN.6 eV UV IR Lyman 10. Hydrogenic atoms only radiate light when they jump from one Bohr-Sommerfeld stable orbit to one of lower energy.50 eV N=2 -3.85 eV -1.`0 hf = λ The spectrum of emitted radiation is determined by the differences in the energy levels.` − EM. 94 . Example Consider a simple quantized atomic system in which an electron can have four energy levels E1 < E2 < E3 < E4 as shown below. hf2.2 = E4 − E2 . to least energetic or reddest) hf4. as opposed to a continuous spectrum in which all colors or wavelengths of light are present.E 4 E 3 2 1 Levels Transitions B R Emission Spectrum These wavelengths are (from most energetic or most blue. hf3.3 = E4 − E3 .2 = E3 − E2 hf4. as in the spectrum of light emitted by 95 .1 = E4 − E1 . and so the line on the film or grating is an image of the source or slit. hf4. hf3. This type of emission spectrum is called discrete line.1 = E3 − E1 .1 = E2 − E1 a better representation of what one would see of this light were passed through either a prism or diffraction grating would be B R Emission Spectrum Keep in mind that we will see spectral lines because the light from the atoms will be passed through a slit. we are very unlikely to find any atoms in energy state E n . a far more correct picture of the atom. f4. This is also called the degeneracy of the energy level E. Basic statistical mechanics can partially explain why some spectral lines are not seen if the gas emitting the light is excited to temperature T . 2 m n2 λ−1 n.6 eV 13. gas with the energy level spectrum of the previous example may not be excited enough to have many atoms in E4 . that certain transitions between energy levels are forbidden on the basis of angular 96 . 2.309 cm−1 What we actually see when we analyze light from thermally excited atoms differs from the predictions of Bohr-Sommerfeld theory only in two details. and so many spectral lines emitted when atoms drop out of high energy states will not be seen unless the gas is hot enough to have an appreciable number of atoms in the requisite high energy starting-states to begin with.1 emitted.6 eV − . Notice that this tells us that if En >> kT for a given temperature. For light emitted by Hydrogen this expression becomes hfn. Example At fairly low temperatures T .2 . we will see no light emissions from atoms falling out of this state. B R Low T Emission Spectrum However we will see in our excursions into wave mechanics. At low T this gas will have emission spectrum like that shown below.m = in which R∞ = mcα2 2h 13. Not all predicted spectral lines (emitted wavelengths) are actually seen. The probability that an given atom in the gas will have its electron in energy level En is En ℘(En ) = D(En ) e− kT in which D(E) is the density of states. Not all emitted spectral lines have the same intensity or brightness.a very hot opaque body or a heated cavity in a conductor (cavity or BlackBody radiation a’la Planck).3 .m == R∞ ( 1 1 − 2) 2 m n is called the Rydberg constant R∞ = 109737. so we will see almost no light of the frequencies f4. If no atoms are to be found in that state. 1. f4. the number of configurations of energy E. some of which are beyond the ability of that theory to explain. the successor-theory to the BohrSommerfeld model. and so in the emission of light by excited atoms.. certain transitions might not be possible because the initial and final states may have angular momenta that differ by the wrong amount for angular momentum conservation. Different spectral lines are of different intensities because of several factors. One can see the basic picture.3 f4. Let us couple this together with the notion that since lignt carries angular momentum 1 h ¯.3 f4. We say that En is (n − 1)−fold degenerate.3 f4. .3 f4. Allow/Forbid forbid forbid allow allow allow forbid forbid allow allow forbid allow allow forbid allow allow allow forbid allow allow .1 f4. and ten transitions of emission 97 . called spin angular momentum.. Suppose that We have four quantum energy levels E1 < E2 < E3 < E4 and they are degenerate. The inability of Bohr-Sommerfeld theory to explain the relative intensity of spectral lines was one of its major failings.1 f4.3 f4.3 f4.momentum conservation. Initial N 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 . there are two allowed transitions that will cause light of frequency f4.3 . 0 We now list some transitions and the frequency of light emitted.1 f4.3 f4.1 to be emitted..1 f4. This fact was not known at the time of Bohr-Sommerfeld theory. one of which is the degeneracies of the two quantum energy levels involved in the transition. and cannot be properly taken into account except within the framework of wave mechanics. Light carries angular momentum of 1h ¯ ..3 f4. the only allowd atomic transitions must obey the selection rule ∆` = ±1. Initial ` Final N 4 1 3 1 2 1 1 1 4 3 4 3 4 3 3 3 3 3 3 3 2 3 3 3 3 3 2 3 2 3 2 3 1 3 1 3 1 3 . and other selection rules.. f f4.. and electron in E2 can have angular momentum 1 h ¯ or 2 h ¯ and so one. An electron in E1 can have angular momentum 1 h ¯ .3 f4.3 f4.3 f4.3 f4. Final ` 1 1 1 1 3 2 1 3 2 1 3 2 1 3 2 1 3 2 1 .3 f4.3 f4. if the temperature is high enough. The theoretical successor. 5.1 = 10 =5 2 and we should see the spectral line of frequency f4. 2.3 and so the ratio of intensities should be If4. The quantity d in the first expression is the distance between slits. R = 25. Gaertner Scientific. Hydrogen light source. 4.3 1. Spectrometer. The figure shows how the setup will look. 3. wave mechanics. Gas discharge tube powered up by a Variac. Type W5MT3.1 line. with the variac connected to the hydrogen discharge tube. 000. Bohr-Somerfeld theory does not give us the correct energy level degeneracies.3 : If4. and so its predictions of spectral line intensities are not correct. the reciprocal of the number of lines per centimeter in the grating. Gaertner Scientific. However it does successfully predict the actual frequencies of the spectral lines to excellent precision. The diffraction grating used in this lab (by me) has m = 1. 10.frequency f4. 000 in which mλ = d sin θ ∆λ R= = mN λ where R is the resolution in first order (m = 1). Diffraction grating. N = 25. 98 . and more applicable to a wider range of problems.3 five times as bright at the f4. more general. and so many of its core ideas and assumptions must have some truth to them. is incomparably more accurate. Apparatus Sodium light source. it is the diffraction order of the resolved spectral line. Remember me c2 = 0.0072974 ≈ 137 e2 with k = 4π0 . Protractor (angle scale) Grating θ H source Slit colimator Telescope with cross hair Next we determine the angles at which maxima occur when light from hydrogen discharge tubes is passed through the grating.511 M eV and hc = 1239. emissions occur with wavelengths hc µc2 α2 1 1 E= =− ( 2 − 2) λ 2 n m mp ≈ me and α = h¯kc is the fine structure in which µ is the electron reduced mass µ = mmee+m p constant 1 α = 0.The first step is to determine the number of lines per centimeter by calibration using the two Sodium lines λ1 = 5889. We then solve for the wavelengths by mλ = d sin θ Do not confuse m with an atomic quantum number. Finally we use the fact that for the Bohr model of hydrogen.8eV nm.923˚ A We will first measure the angles at which the lines will be scattered by the grating. with = R∞ ( 1 1 − 2) 2 ni nf 99 . There will only be a few visible wavelengths from which we will deduce 2π 2 µk 2 e4 1 1 1 = ( − ) 2 λ h3 c ni n2f for ni → nf principle quantum number transitions. λ2 = 5895.953˚ A. 1889 28.5500 24.6167 average (deg) 40.2000 28.7167 25.the Rydberg constant R∞ .750 25.2o 36.1833o 36.299 ± 558.233o Of course when you actually perform the experiment.99˚ A sin(36. What you should expect to see as you move the collimator from angle to angle is something like the following m=-2 m=-1 m=0 m=1 m=2 θ 0 10.1833 28.233o Trial 3 36.59˚ A sin(36.6333 100 trial 3 (deg) 40. color quantum number red n=3 green n=4 blue 1 n=5 blue 2 n=6 trial 1 (deg) 40.5556 24.1833o .76˚ A The Rydberg constant from Hydrogen spectra With ni = 2 and nf = n.232o from which we obtain the averages θ¯1 = 36.5500 24.923˚ A = 9974.7167 25. you will need to do standard error analysis. θ¯2 = 36.7278 25.4 Data and analysis λ 1 2 Trial 1 36.5500 trial 2 (deg) 40.233o Trial 2 36.953˚ A = 9975.233o ) and for an average linespacing of a ¯ = 9975.1833 28.1833o 36.183o ) a2 = 5895.6000 . These give line spacings for the grating of a1 = 5889.5667 24. 20 0.0/λ R∞ = 0.24 which is in good agreement ( error is 4%) with the accepted value of .16 ˚ A = 643. to within 2%. 0.22 (n2−4)/4n2 0.22222 1 22 n 3 4 5 6 1 λ . These lines are transitions from the m = 2 level into one with quantum number in the table.52 ˚ A You can compare these with the exact.000155348 .22 1000.26 . and doing a linear regression.38 nm. − n12 .18 0.13889 . known values reported in the next experiment.001057 ± 0.18750 . λ5 = 4303.000208601 . λ6 = 4152.7 nm.82 ˚ A = 479. agreement with known wavelengths is very good.000232382 .16 0.14 0.14 0.24 ˚ A.12 0.18 0.From this data and a ¯ we find that λ3 = 6437.16 0.2100 .20 0.000240817 The average value for the Rydberg constant from these four data can be gotten by plotting the reciprocal wavelength versus the difference in the squared reciprocal quantum number difference.24 0.00109737A−1 101 0.00010792A−1 0. λ4 = 4793. where r0 is the distance from point P to the central slit.10. and d is the slit seperation. very fine grooves are cut into a transparent glass or polymer slide. These grooves act like aperatures and light transmitted through them undergoes multiple-source interference. We represent a light wave of polarization Pˆ and wavelength λ as 2π 1 E0 2π ~ = pˆ√ E sin ( r − ωt) = pˆ√ Im ei( λ r−ωt) r λ r and notice that if we label the central slit j = 0. then the j th slit is approximately rj = r0 + jd sin θ away from point P . ~T = E ≈ N X 2π E0 pˆj √ Im ei( λ rj −ωt) rj j=−N N X 2π E0 pˆ0 √ Im ei( λ rj −ωt) r0 j=−N 102 . From the figure below representing the grooves or cuts seen edge-on d r 1 r P 0 θ θ r-1 λ we can use superposition to construct the intensity I at the point P on a screen upon which the transmitted light will fall.5 Diffraction gratings Diffraction gratings are essentially multiple-slit aperatures that can be either reflection or transmission devices. We can add up all of the waves arriving from all slits at point P . In a transmission grating. As N increases the disparity in relative brightness grows. 103 .N X E0 2π r0 −ωt) i( 2π λ √ ≈ pˆ0 Im e ei λ jd sin θ r0 j=−N This sum is not difficult. Notice that the bright fringes or maxima have an amplitude of about N 2 times the brightness of the N − 2 secondary maxima that lie between consecutive primary fringes. call 2π x = ei λ d sin θ then it is N X j=−N xj = x−N (1 + x + x2 + · · · + xN ) = x−N 1 = = 1 − xN +1 1−x 1 xN + 2 − x−(N + 2 ) 1 1 x 2 − x− 2 θ sin ((2N + 1) πd sin ) λ πd sin θ sin ( λ ) and we can assemble the intensity from the time averaged Poynting vector I(θ) = θ θ ) 2 ) 2 sin ((2N + 1) πd sin 1 E02 sin ((2N + 1) πd sin λ λ ( ) = I ( ) 0 πd sin θ πd sin θ 2µ0 c r0 sin ( λ ) sin ( λ ) In this case our grating has 2N + 1 slits. the corresponding expression for an N -slit grating is sin ( N πdλsin θ ) 2 I = I0 ( θ ) sin ( πd sin ) λ For N = 11 the intensity of transmitted light versus angle looks like the following figure. bright primary maxima with complete darkness between them as N grows very large. until we see only very sharp. 11 slit 140 120 100 I(θ) 80 60 40 20 0 −2 −1 0 (πd sinθ)/λ 1 2 Typical gratings will be quoted as having a certain number of rulings or grooves cut into them per centimeter.0 × 10−2 m = 1.0 × 10−6 m = 1000. 000 rulings per centimeter has a d-value or slit-to-slit seperation of d= 1. Once d is known. One must first establish the value of d for the grating being used by measuring the angular position of a well known or standard wavelength of light. These will not superimpose perfectly.0 nm 4 10 If we pass a light sample containing several distinct wavelengths of light through the grating. For example a grating with 10. we will see one diffration pattern per wavelength present in the sample. and by measuring the spread we can determine the actual wavelengths. such as that from a laser or from a mercury lamp. This is what is done in the Rydberg constant experiment. the light to be measured is passed through the grating 104 .Diffraction grating. 236 105 . degrees 10 15 We see two fringes resolved in first order since the m = 1 interference fringes for each color are well separated. Carefully measuring with a ruler we see that the m = 1 fringes have angular positions λ λ1 λ2 θ (degrees) 7. Consider a spectrum containing two colors of visible light passed through a diffraction grating.and the angular positions of the various wavelengths present is measured. producing the intensity pattern seen below Resolution in first order 300 250 I(θ) 200 150 100 50 0 −15 −10 −5 0 5 θ.697 9. 0×10λ2 nm yellow-1 579.697) = 133.138 blue-violet 435.0 4.232 violet 434.322 blue-green 486.943 106 faint faint faint faint faint faint faint .004 green 546.065 blue-2 491.309 deep-violet 410.983 yellow-2 577. nm 1.265 deep-violet 407.353 blue-1 496.6 4. or through a prism.9 5.265 blue-violet 433.106 Hydrogen spectral lines red 656. This is done by creating a calibration curve for the instrument using a known sample. The idea is that we use the instrument to measure the angular location of particular spectral lines passed through the grating. and match these lines to known wavelengths.8 6.0 nm Then since the positions of the primary maxima of the grating are d sin θ = mλ we discover that λ1 = (1000.013 deep-violet 404.7 5.0 2. and so d = 1000.265 blue-violet 434.7 6.5 nm 11 Measurement of unknown spectral lines We can use the same apparatus to identify an unknown sample of gas from its visible emission spectrum. Mercury spectral lines 6 −2 Color λ. and found that it had 10. but this will be nonlinear. 000 rulings per centimeter.Supose that we have previously measured our diffaction grating’s d value.0 nm) sin(9.9 nm and λ2 = (1000.0 nm) sin(7. We can make a plot of λ versus angle θ.0 3.1 4.2 5.1 3. so instead we will plot the sine of the angular displacement of a line versus λ.0 5.3 2.8 5. the best being Mercury due to the number of visible spectral lines.236) = 160. 45 203.3 192.8 192.383 151.5 10.333 192.2583 20.1 The Mercury data Using the same apparatus as in the previous experiment.225 31.1333 20.6 202.225 19. n=2.867 163. we obtain a ¯ = 33800 ˚ A.3o .0333 29.883 172.667 159. L n=1.8333o .25 Average Displacement θ of diffraction 7. Instead we will correlate these lines with true wavelengths to produce calibration curves relating sin θ to λ for each diffraction order n.55 9.75 n = 2 first order 198 167.0417 10.367 213. 107 .667 153.11.0083 31. R n=2.4 n = 3 first order 205.017 162. 175o 180 = 175. we calibrate the grating spectrometer with a Mercury source. and so forth.0833 of diffraction 15.4 175.217 213.483 151.917 172. using the four brightest lines. If we were to do it now for this data and this grating.967 162. R θ displacement 182 50’ I found the instrument to be centered at 182o 500 = 182.L n=1.3083 of diffraction 23.333 173.55 201. and the angular positions and displacements for the first three diffraction orders to be Line violet green yellow-2 yellow-1 violet green yellow-2 yellow-1 violet green yellow-2 yellow-1 Left Right n = 1 first order 190.6 211.4o . for example 190o 240 = 190. which we can use to determine the wavelength of an unknown spectral line from its angular displacement. We now could compute the diffraction grating spacing a from nλ = a sin θ but we have already done such a thing in the previous experiment.1167 In this table all angles have been converted to decimal degrees from degrees and minutes. 00 nm.222900 sin θ with a χ2 = 0.222900 sin(0. right angular position. which is a very good fit.74 nm which is in superb agreement with the known red-Sodium line at λtrue = 615.0 579.17508 A linear regression using the program lg in the support software or from the 499 website archive on rustam.edu on this λ versus sin θ results in a straight line λ = 4.0 Average Displacement θ rad 0. with an average displacement of θ = 10.8 546. We determine its wavelength to be λ = 4. 194o 140 with the instrument centered on 183o 300 .16581 0. 173o 140 and right. left.uwp.18733) = 615. 11.11.18733 rad.655966 + 3281.655966 + 3281.17599 sin θ 0. and we measure this lines wavelength from the calibration curve to be λ = 4.17436 0.7333o = 0.13139 0.43 nm! The yellow Sodium line is seen in first order n = 1 diffraction at angular positions right. This results in an average angular displacement of θ = 10.13177 0. 108 .16505 0.2 First order calibration curve Line violet green yellow-1 yellow-2 λtrue nm 435. red.179333) = 589.179333 rad.275o = 0.655966 + 3281.17526 0.1 577.222900 sin(0.94 nm again in very good agreement with the true value λtrue = 589.000151 for the fit. 172o 460 .3 Measuring Sodium lines The Sodium source produces the following lines in the first order set of diffraction maxima. 193o 470 . 67 × 10−8 12. Z S 0 E · n dS = We now apply the well known vector calculus identity I C=∂S V · d` = Z 109 ∇ × V · n dS I C=∂S B · d` . Electromagnetic portion. is a perfect absorber since once light enters the hole its odds of bouncing back out are slim. the proportion emitted at various wavelengths could not be successfully predicted by classical physics. We begin this investigation with its electromagnetic component. You are familiar with Maxwell’s equations in integral form from elementary physics I S E · n dS = qinS . the currents of moving charges.1 W m2 K 4 Background If a metallic box enclosing a cavity is slowly heated . σ is the Stefan-Boltzmann constant σ = 5. The whole problem of heat radiation is complicated by the fact that it lies at the intersection of several very distinct fields of physics. any light that enters will be absorbed and re-emitted by the walls. unheated.12 Stefan-Boltzmann Radiation Law The purpose is to verify the radiated power versus temperature for a blackbody radiator. µ0 IthruC d + µ0 dt Z S 0 E · n dS = I C=∂S B · d` In empty space both the static charge q and conduction current I. computation of the fraction of energy radiated within a given wavelength range.and the wavelengths of light emitted through a small hole in one wall is monitored. The box is a perfect absorber since if it’s temperature is stable. since light that enters the hole will be absorbed and re-emitted. A perfect absorber is also a perfect emitter (Kirchoff’s law). This is our first basic problem. leaving Faraday’s and Ampere’s laws in a highly symmetrical form d dt Z S B · n dS = − I C=∂S d µ0 dt E · d`. The box itself. 0 I S B · n dS = 0 and d dt Z S B · n dS = − I C=∂S E · d`. in particular electromagnetism and thermodynamics. and ultimately this energy must escape or the box temperature would rise. ℘=A Z 2πhc2 ∞ 0 hc λ5 (e λkT − 1) dλ = σAT 4 in which A is the surface area of the radiator. are zero. We will now take for granted that light is an electromagnetic wave. in empty space ∇ · E = 0. first to determine the spectrum of frequencies of standing ling waves that can exist inside the box. By the time of Planck’s heat radiation experiments. to determine how many distinct waves have the same frequency. To do this. and contemplate a hollow metallic box containing electromagnetic standing waves. without being annihilated by interference. or how many distinct waves have frequencies between f and f + df . with normal n and right-handed bounding curve C. a fact confirmed experimentally by Hertz. We apply this to the two static equations (Gauss’s laws) to obtain. µ 0 0 E = ∇ × B dt dt To proceed further we apply another vector calculus identity I S=∂V V · n dS = Z V ∇ · V d3 x in which S is a closed surface enclosing volume V . it was widely accepted that radiant heat was a form of electromagnetic radiation with long wavelenghts compared to visible light. µ 0 0 2 E = ∇ × B dt dt dt and substitute one into another to arrive at −∇ × (∇ × E) = µ0 0 d2 E dt2 and now use a third vector identity to simplifiy the cross product −∇ × (∇ × E) = −(∇(∇ · E) − ∇2 E) = µ0 0 d2 E dt2 and use the differential form of Gauss’s law to arrive at the electromagnetic wave equation 1 ∂2 2 (∇ − 2 2 )E = 0 c ∂t in which the speed of propagation of the waves so constructed is c= √ m 1 = 3 × 108  0 µ0 s the previously established speed of light. first differentiate the two dynamical equations to get d d2 d ∇ × B = −∇ × (∇ × E). This allows us to write both sides of the two dynamical laws of electromagnetism as d d B = −∇ × E. in thermal equilibrium with the hot walls. We will need to do this in order to compute the 110 .in which S is a two-dimensional surface in space. ∇·B =0 Maxwell used these four equations to prove that light is an electromagnetic wave. and second. We now have two tasks. 10 5 0 −5 −10 −10 −5 0 5 10 In two dimensions a light wave propagating an a given direction can have only one possible polarization direction perpendicular to its flow of energy. m) becomes the area of a quarter circle. m. x = a. y = 0.spectrum of light emissions from a small hole cut in the wall. y = a. so that no electric field can penetrate into them. √ For example in two dimensions let N (ω) equal the number of modes in a radius of r = n2 + m2 around zero frequency. z = a. p. so π 1 N (ω) = 1 · πr 2 = 2 a2 f 2 4 c The number of modes per unit frequency interval is then dN (f ) a2 = 4πf 2 df c In three dimensions we find that N is twice ( because any light wave propagating in direction n can have two independent polarizations p such that p · n = 0 in three dimensions) 111 . Then as r becomes very large the number of combinations(n. If the walls of the box are for example perfect conductors. then we must have E = 0 on the walls we find the solution Ex = E0 sin( mπy pπz iωt nπx ) sin( ) sin( )e a a a for a cubic box of side a such that (n2 + m2 + p2 ) ω2 π2 = a2 c2 and each mode in the box is specified by these three integers n. and z = 0. and if the walls are at x = 0. The next stage is to compute the total energy possessed by the radiation within the box according to Z 2¯ X dN dU U d U (f. as Planck was to show. the box would contain an infinite total energy. This requires input from thermodynamics to compute the average energy that a light wave. The average energy stored as electromagnetic waves in the box per volume per frequency interval is dN (f ) 8πkT f 2 d2 U (f ) = u¯(f ) = df dV dV c3 This is the Rayleigh-Jeans law. since it contains ever increasing amounts of energy at higher and higher frequencies. Thermodynamic portion. since in classical electrodynamics the energy density of electromagnetic radiation is proportional to the electric field amplitude squared.one-eigth the volume of a sphere of radius r and so 1 4 3 8πa3 f 3 N (f ) = 2 πr = 83 3c3 so that dN 8πa3 f 2 = df c3 This is referred to as the density of states for electromagnetic waves in the box. 112 . According to the classical equipartition of energy each quadratic when in equilibrium with degree of freedom in the Hamiltonian has average energy of kT 2 a heat bath. This law predicts an ultraviolet catastrophy. which we will regard as an oscillator since it satisfies an oscillator equation. can extract from the walls at equilibrium. and sum over all of the modes. It is of course not in agreement with experiment. Using the Gibbs Canonical Ensemble. ¯= E Z ∞ 0 ℘(E) E dE = E 2π E e− kT dE ω R ∞ 2π − E kT dE 0 ω e R∞ 0 = kT This amount of energy then determines the amplitude of the wave. T ) = 3 = ·u ¯(f ) = df dV a dV df f df in other words we multiply the number of modes of frequency f by the average energy of such a mode when in equilibrium with the hot walls at T . since the sums are geometric P∞ hf −n kT hf ne E¯ = P∞n=0 hf −n kT n=0 e hf = hf kT hf − (1−e kT )2 e 1 1−e 113 − − hf kT = hf e hf kT −1 . 2. This cannot be true. with an average determined by the temperature.. (the Planck hypothesis) and the constant h needs to be determined by fitting the experimental data. which is in fact an oscillator by virtue of being waves. 1.. We find now that E¯ = Z ∞ 0 ℘(E) E dE = E 2π E e− kT dE ω R ∞ 2π − E kT dE 0 ω e R∞ 0 → n −E kT n=0 En e P∞ n −E kT n=0 e P∞ Fortunately this is no harder to compute than the integrals of the Equipartition version. This assumption of discrete oscillator energies as opposed to continuous simply translates into the need to sum rather than integrate in order to compute the average energy per mode. Planck noticed that the experimental data can be fit with tremendous accuracy by instead assuming that each mode. or the theory is doomed to ultraviolet catastrophe.T)/df dV 600 400 200 0 0 5 10 hf/kT 15 20 The problem lies in the assumption that the equipartition law is valid. 2.can only have a discrete spectrum of excitation energies rather than all possible energies. 1.5 T0.Rayleigh−Jeans curves for T=T0. He proposed that the energy of each mode is n (hf ) in which n is 0. there must be dependence upon the frequency.0 T0 800 d2U(f. T)/df dV 10 8 6 4 2 0 0 5 10 hf/kT 15 20 hc k Which can be found by maximizing the energy per frequency interval with respect to frequency. assuming that the dominant constituent of the universe is the electromagnetic field filling it.0 T0 12 d2U(f.2012 The Stephan-Boltzmann law has been used to estimate the size of the universe as it cools.the Stefan-Boltzman law dU = dV Z ∞ 0 8πhk 4 T 4 d2 U df = df dV c 3 h4 Z ∞ 0 8πk 4 π 4 4 4 x3 dx = T = σT 4 ex − 1 c3 h3 15 c 2 U versus in which σ = 5. 2. The graph of dfd dV f has a maximum value at a wavelength predicted by the Wein law of displacement Planck curves for T=T0.67 × 10−8 m2WK 4 is the Stefan-Boltzmann constant. since cooling must be at constant total energy. 1. and correctly predicts the total energy stored in the box as a function of temperature. and inserting this into the expression for the energy density per frequency interval in the box yields dN (f ) 8πf 2 hf d2 U (f ) = u ¯(f ) = 3 hf df dV dV c e kT − 1 which is in exact agreement with the experimental data. From U = 4c σ V T 4 we obtain the expansion rule V1 T14 = V2 T24 114 .This is the celebrated Bose-Einstein distribution function.5 T0. λmax T = 0. The box is a distance r < c dt away from the hole of area A. all moving isotropically away from the box at speed c. As a function of the temperature of the box.12. hf 3 df c (e kT − 1) 8πf 2 kT 2 r dr sin θ dθ dφ (Rayleigh-Jeans) c3 all in the form of photons.2 Luminosity or radiance Consider now the problem of the luminosity of a small hole cut into the wall of the cavity. dV c3 (e kT − 1) 8πf 2 kT (Rayleigh-Jeans) c3 and so the little volume element contains energy within freqency interval [f. The energy per unit volume in the cavity is 8πhf 3 d2 U = (Planck). f + df ] of dU 8πhf 3 = r 2 dr sin θ dθ dφ (Planck). c dt=R dr θ dV r θ n -z The fraction of the radiant energy in the volume element heading in a direction n that will take it through the hole is Az·n angle subtended by hole A cos θ 2 = r = 4π 4π 4πr 2 and so the energy from this particular volume element that will leave through the hole within time dt is A cos θ 8πhf 3 8πf 2 kT 2 A cos θ d∆Uthru r 2 dr sin θ dθ dφ = (Planck). r dr sin θ dθ dφ (Rayleigh-Jeans hf 2 3 df 4πr c 4πr 2 c3 (e kT − 1) 115 . Consider a volume element dV = r 2 dr sin θ dθ dφ within the cavity containing electromagnetic field energy. what is the total flow of power in the form of heat radiation through the hole? In the figure below we see a hole of area A. hf df. 3 Subtronics model 2000 multi-meters.We now add up the contributions from all of the volume elements within a hemisphere of radius R = c dt against the wall of the box containing the hole. Stefan-Boltzmann Lamp. This is true for all Blackbody radiators.total 2 2 = = r dr sin θ dθ dφ hf hf df 4πr 2 c3 (e kT − 1) 0 0 0 c3 (e kT − 1) 4 and for Rayleigh-Jeans π 8πf 2 kT Z R Z 2 Z 2π 2 8πf 2 kT c dt A A cos θ d∆Uthru. Pasco Scientific model TD-8555. Power supply. 3. 116 .3 Apparatus 1. Micronta dual tracking DC power supply. 0-20 V. such as the metal filament used in this experiment. Thermal radiation sensor (thermopile). 4. Pasco model TD-8553. 12.total = = r dr sin θ dθ dφ df c3 4πr 2 c3 4 0 0 0 Now divide by dt and use P = radiated through the hole dU dt to get the power at frequencies between f and f + df 8πhf 3 c A dP = Planck hf df c3 (e kT − 1) 4 dP 8πf 2 kT c A = Rayleigh-Jeans df c3 4 The ultraviolet catastrophe is now pretty obvious for Rayleigh-Jeans. integration over all frequencies results in an infinite power flow for the Rayleigh-Jeans law. 2. but results in the well-known and correct Stephan-Boltzmann law for the Planck version Ptotal = Z ∞ 0 Z ∞ 8π(kT )4 π 4 c A 8πhf 3 c A dP df = df = = σ AT4 hf 3 h3 3 df 4 c 15 4 0 kT c (e − 1) for the power flowing from the hole. for the Planck case Z R Z π Z 2π 8πhf 3 8πhf 3 c dt A A cos θ d∆Uthru. to be used as voltmeters and an ammeter. The apparatus is set up as in the figure. The resistance R= V I 117 . and I through the filament are recorded along with the detector voltage.0µm will be monitored by placing the lamp a fixed distance from the detector. The power received by the heat sensor which detects thermal radiation in the range 0.5µm ≤ λ ≤ 25. The lamp is powered up and the voltage and current V . and that the current through the lamp never exceed 3 A. Ammeter 0-20 V Power Model TD-8555 Stefan-Boltzmann Lamp CAUTION Voltmeter Voltmeter It is very important that the voltage applied to the lamp never exceed 13 V. The first step in determining the temperature function is to measure the resistance of the filament at room temperature.of the filament is a well known function of it’s temperature. hopefully around 300K. provided by the manufacturer. and at high temperatures scales in a very precise way with respect to a reference resistance. 118 . its resistance will be related to the temperature by finding the ratio R(T ) R300 and looking up the corresponding temperature in Kelvin in the table below. To do this. and measure the resistance R300 of the filament with a multi-meter. When the filament is hot. carefully measure the room temperature. T = R − Rref + Tref αRref at low temperatures. 95 17.86 9.58 7. made from this data.29 16.28 8.03 6.63 16.97 19.34 2.14 7.44 10.R R300 1.99 15.72 14.63 11.34 14.66 Temperature (K) 300 400 500 600 700 800 900 1000 1100 1200 1300 1400 1500 1600 1700 1800 1900 2000 2100 2200 2300 2400 2500 2600 2700 2800 2900 3000 3100 3200 3300 3400 3500 We could also interpolate the graph below.36 3.87 2.62 18.24 11.03 10.41 4.95 5.88 4.28 18.0 1.43 1. 119 .84 12.71 8.46 13.85 3.08 13.48 6. 1500 1000 500 0 0 5 10 R/R_300 15 20 Instead we will use a Lagrange interpolation program that has this table coded into it. R[1]=1.T[34]. float r.alpha.87.T[2]=500.c1.0.c3.T[3]=600.h> #include<stdlib. main(int argc. char *argv[]){ R[0]=1. It uses T (R[n] + δr) = − (δr − 1)(δr − 2)(δr − 3) δr(δr − 2)(δr − 3) T [n] + T [n + 1] 6 2 δr(δr − 1)(δr − 3) δr(δr − 1)(δr − 2) T [n + 2] + T [n + 3] 2 2 and is given below − /* temperature interpolator for SB experiment */ #include<stdio.0.0.h> double R[34]. R[2]=1.T[1]=400.0. R[3]=2. int n.43.c4.34.T[0]=300. 120 .c2.m.0.y.h> #include<math.3500 3000 2500 2000 Temp. c3=-alpha*(alpha-1. R[8]=4.T[12]=1500. y=c1*T[n]+c2*T[n+1]+c3*T[n+2]+c4*T[n+3].44.0)/2.0. R[17]=10.T[18]=2100.T[20]=2300. for(m=0.48.T[30]=3300.0.36. /* find R in table closest to input r */ n=0. R[32]=19.85. R[29]=17. R[7]=4. R[20]=11.58.T[11]=1400.0.T[8]=1100. R[18]=10. R[27]=16. R[21]=12.0)/6. c2=alpha*(alpha-2.99. c4=alpha*(alpha-1.0)*(alpha-3.0.T[14]=1700.T[23]=2600.0.0. R[6]=3.84.0. R[16]=9.0.0.14.24. R[28]=16.0. R[23]=13. R[30]=18.63. R[25]=14.0. R[19]=11.T[17]=2000.T[19]=2200.0.T[4]=700.29.62.0.y).0. R[31]=18.T[6]=900. } /* now we have bracketting values */ alpha=r-R[n].T[29]=3200.T[21]=2400. R[5]=3. c1=-(alpha-1. } 121 .0.0)*(alpha-2.T[10]=1300. R[12]=7.63. R[15]=8.41.28.0)*(alpha-3.0. printf("R/R(300) = %f at T = %f\n".0.34.T[16]=1900.0)/6.T[32]=3500. R[14]=8.T[13]=1600.0.T[26]=2900.0.0.0.28.T[9]=1200.03.0.r.T[22]=2500.T[31]=3400. R[26]=15.T[24]=2700.0)*(alpha-2.0.T[28]=3100.0.71. R[10]=6. R[11]=6.0.0. r=atof(argv[1]).97. R[22]=13.m++){ if(fabs(r-R[m]) <= fabs(r-R[n])) n=m.T[25]=2800. R[9]=5. R[13]=7.88.95.m<31.T[5]=800.0.0.0.0.0.T[27]=3000.66.46.0)/2.86.08.R[4]=2.0)*(alpha-3.0.72.03.T[7]=1000.95.0. R[24]=14.T[15]=1800. 0005 Sns V (mvolts) 0.We simply run it by passing it the RR300 that we want T for. wait until all meter readings stabilize.36 1.99 31.26 5. " ". $2.867 2. $3-0.56 Dark Background Fil.08 6.28 21.0005. 12. "&".005 0.289 0. 12. " ". $2-0. I (amps) 0.001 Sns V (mvolts) -0.043 0.03}’ data>newdata and the resistances found with awk ’{print $1.10 Fil.5 Data and analysis In 1996 I took the following data for this experiment.57 4.830 4.88 2. Fil.03 This data is corrected with the awk script awk ’{print $1-0.21 2.09 2. $3. apply a voltage to the lamp. "&".144 0. This background must be subtracted from all of the subsequent detector readings.001. "\\\\". I (amps) 0.72 1.99 7.24 Fil. which is proportional to the power it receives within the given wavelength band.29 6. You will find this program as a download on the 499 website archive.95 11. and record filament current.78 25.4 Procedure First measure the background reading of the thermal sensor with no power applied to the lamp. "&". V (volts) 0. voltage and detector voltage.220 0. V (volts) 0. $1/$2. "\hline"}’ newdata which results in the table 122 .636 0.36 15. With the lamp and detector a fixed distance apart.933 1. 719 1.8 4.0005V = 0.5µm 2πhc2 hc λ5 (e λkT − 1) dλ = σAT 4 which we compute for each temperature using the following C program #include <stdio.25 21. /* for looping */ double lambda.95579 5.89073 2.02154 6.96 31.2595 5.5695 4.2895 6.4066 6. I (amp) 0.01077 3.8 4.19 0.089 2. 123 . The total power radiated within the wavelength band sensed by the IR detector is ℘=A Z 25. $1/(0.879 2.0µm 0.929721 1.Fil.33 15. "\\\\".75 25.209 2.h> int n.239 Sns.5*$2). integral. the corrected voltages and currents.07 R(T ) R300 1.454331 0. "&".92 11.9895 7.3282 6.07 R (ohm) 0.5595 Fil. constant2.1641 3. V (volt) 0.8665 2. with awk ’{ print $3.932 1.factor.5Ω R300 = 0. V (mvolt) 0.33 15.2885 0.75257 T (K) 500 875 1100 1175 1300 1360 1430 We now compare our results to the theoretical predictions. "\\hline"}’newdata which gives us the table Sns.7033 3. dlambda.47789 2.635 0.25 21. constant1.96 31.h> #include <math. V (mvolt) 0.0795 6.75 25.001A and we then process our data file ”newdata”.359 1.85944 3.h> #include <stdlib.92 11. 0. essentially the ”dark” filament line of data above.37628 The resistance at room temperature is gotten by measuring the current and resistance through the filament at very low applied voltage.78146 4. since by hypothesis they are proportional to one another. T.} T=(double)(atof(argv[1])). integral=0.55 32695.float T.96 1360 31. } integral=integral*dlambda*constant1. dlambda=0.8 82104. exit(0). factor=lambda*lambda.0/factor. however the sensor voltage is proportional to the received power. char *argv[]){ if(argc < 2 || argc >2){ printf(‘‘ .25 1175 21.0E-7 */ constant2=0. and so the plot of ln Vsens versus ln ℘ should be a straight line of slope one. n<=2500.749176e-16. /* constant1 = 2 pi h c^2 */ /* constant2 = hc/k */ main(int argc. factor=factor*lambda. constant1=3.75 1300 25. printf(‘‘%f\t%f\n’’. factor=factor*factor. for(n=50. /* lambda^5 */ factor=factor*(exp(constant2/(lambda*T))-1.33 1100 15. } Our sensor does not recieve all of the radiated power./sb temp’’\n). but a fixed fraction depending on it’s surface area and the distance from the filament.8 500 4.5 107002 160530 192369 235243 The log-log plot of the theoretical power versus sensor voltage is below 124 .0).014413. The C program gives the data Vsense (mV) T (K) 0. integral). /* dlambda=1.92 875 11. integral=integral+1.0.0000001.07 1430 ℘theory 3350.n++){ lambda=(float)n*dlambda. 106 105 logP 104 103 0.1 1 10 100 logV_sens which has a slope of 0. 125 . confirming within an accuracy of 7% the Stefan-Boltzmann Radiation law.933. 0. 0. 0. p2 c2 + m2 c4 ) ( . 0. In the Compton effect a stationary electron is bombarded with photons and the scattering angles and energies satisfy standard relativistic versions of energy and momentum conservation laws. sin φ. there is evidence that in collisions with matter light behaves like any other particle. γ m φ γ θ m The relativistic dispersion relation E= q p 2 c2 + m 2 c4 suggests that for massless particles such as light the magnitude of the momentum vector is E/c. 0. The De Broglie relation E hc p= = c λ allows us to write the conservation of energy-momentum equation p˜bef ore = p˜af ter q hf 0 hf 0 hf 2 0 cos φ. hf )+(0.1 Background The concept that light can be treated like a particle in some applications is not off the wall. mc ) = ( c c c This gives us two momentum conservation hf hf 0 = cos φ + p cos θ. 13. c c 0= hf 0 sin φ − p sin θ c and one energy conservation law hf + mc2 = hf 0 + 126 q p 2 c2 + m 2 c4 . −p sin θ. hf )+(p cos θ.13 Measurement of the Compton Edge The purpose of this experiment is to test the validity of the Compton Scattering expression for γ rays scattered through large angles (180o ). 0. h ff0 c f mc2 1 1 ( 0 − ) = (1 − cos φ) h f f = λ to arrive at the Compton scattering formula ∆λ = λ0 − λ = h (1 − cos φ) mc in which φ is the angle that the light is scattered through. equate them (hf )2 + (hf 0 )2 − 2h2 f f 0 = (hf − hf 0 )2 + 2mc2 (hf − hf 0 ) reduce them algebraically to 2mc2 (hf − hf 0 ) = 2h2 f f 0 (1 − cos φ) or and use mc2 f − f 0 ( ) = (1 − cos φ). This equation makes a prediction about what direction light will be scattered into when it is involved in a collision with static electrons. From our conservation law for energy we see that h(f − f 0 ) = q p2 c2 + m2 c4 − mc2 = Ek which is exactly the kinetic energy of the electron. 1 h mc2 + hf (1 − cos φ) 1 = + (1 − cos φ) = f0 f mc f mc2 127 . and therefore can be verified (or invalidated) by experimentation. c c ( hf 0 sin φ)2 = p2 sin2 θ c hf 0 hf 0 hf hf 0 2 2h2 f f 0 hf − cos φ)2 + ( sin φ)2 = ( )2 + ( ) − cos φ c c c c c c2 or p2 c2 = (hf )2 + (hf 0 )2 − 2h2 f f 0 cos φ We have two equations for p2 c2 .Take the last relation and square it (hf + mc2 − hf 0 )2 = p2 c2 + m2 c4 = (hf − hf 0 )2 + 2mc2 (hf − hf 0 ) + m2 c4 and square the first two relations and add them ( to get p2 = ( hf hf 0 − cos φ)2 = p2 cos2 θ. But we also know that mc2 1 1 ( 0 − ) = (1 − cos φ) h f f which we can solve for f 0 . The detector produces electronic pulses of intensity proportional to the electrical energy dissipated in the crystal. A common version of this experiment is the measurement of the Compton Edge at which the photon is backscattered at angle θ = π. and measure the stopping voltage needed to keep post-collision electrons from arriving at a collection plate. Another signal. Using a series of photoelectric dynodes. the scintillator produces a signal (photopeak) due to the primary radiation. liberating electrons. The channel ch of the MCA output is linearly related to the energy of the event that caused the signal ch = a + b E 128 . by using it to photoelectrically eject electrons.max = hf 1 2 1 + mc 2hf These signals are sorted by the MCA. The detector produces pulses proportional to the energy carried by these electrons. All experiments indicate that this result. which of course can acquire any energy from the primary gamma ray.max = hf 2 hf = hf 1 + 2 mc2 1 + mc 2hf This is the maximal kinetic energy that a stationary electron can obtain in a collision with a photon. gotten by assuming that light behaves like a particle in its interactions with matter. φ = π for which 1−cos φ = 2. is correct.or f0 = f mc2 = mc2 + hf (1 − cos φ) 1+ f hf (1 mc2 The electron kinetic energy is then Ek = hf − = hf 1 1+ − cos φ) hf − cos φ) hf (1 mc2 hf (1 − cos φ) mc2 hf + mc2 (1 − cos φ) which is clearly maximized by a direct back-scattering of the light. where they will complete a circuit and create a current in the apparatus. It is easy to set up the experiment. and is called the Compton edge energy Ee . and a multichannel analyzer. is due to primary radiation scattering electrons out of the conduction band of the crystal. a scintillation detector. From our work above we see that this results in photons of energy hf Eγ Eγ0 = hf 0 = 2hf = γ 1 + mc2 1 + 2E mc2 The actual experiment is carried out with a γ-ray source. producing primary radiation of energy Eγ . which records the number of signals recieved versus intensity or channel. Gamma rays entering the detector undergo Compton scattering from electrons in a crystal (NaI doped with Thallium). the Compton Plateau. giving us hf 2 mc 1 2 Ek. right up to the maximal value at the Compton edge 0 ≤ Ek ≤ Ek. and so the device can be calibrated by recording the channels of two gamma ray sources of known energy.2 a= A typical MCA output looks like the following.1 − ch1 Eγ. Co sample.2 MCA output 2500 Counts 2000 1500 1000 500 0 0 200 400 Channel ch 600 800 The ideal output would be a sharp spike at a channel corresponding to the primary gamma ray Eγ . The actual Compton edge occurs about half-way up the plateau (at ch ≈ 330 on the figure). 129 .max .2 from which b= ch1 − ch2 . ch2 = a + bEγ. and a sharp-edged plateau extending from 0 to Ek. 2. 13.1 − Eγ. Eγ. ch1 = a + bEγ.2 Eγ.2 1.1 . Apparatus 137 60 Cs sample. ch2 Eγ.1 − Eγ. but these features are broadened by secondary scattering and other phenomena. so electrons can be promoted into the conduction bands much more easily. PHA M N x10 4 N Dead Time % count 2 1024 TL cursor M 512 MCS M+1 Nx10 3 4 2 1/2 2/2 5 6 1/1 com 0 ADC offset vert. the energy levels spread out to form energy bands as shown below. Power supply. in a metal this gives rise to electrical conductivity. and be very delocalized in the crystal. so it is very hard to promote a valence electron into a conduction band level. threshhold channel Power on Intensity Roi Expand Scan on off roll Enter off Clear on off on Gate HV Signal off 120 VAC Scintillation counter Lead bricks Power supply 120 VAC 13. spin down). 0-6 mA.3 Scintillation counters and photomultipliers When a large collection of atoms in a gas or liquid are condensed to form a solid (crystalline). two electrons per band level (spin up. 6. 5. NaI Scintillation detector. Conduction electrons will occupy the upper energy conduction bands. 4. Canberra Series 30. vertical 2048 ADC gain 1024 512 256 2048 X1 ACQ mcs pha mcsr add sub col time memory Amp gain X2 I/O out in sel out tty ext I/O LLD adc zero ULD high low input lev. and there is an appreciable gap in energy between valence and conduction bands. Valence electrons will fill the lower energy valence bands. 61516 A DC power supply.3. In semi-conductors this band gap is very much smaller. Lead bricks and sheeting. 130 . In an insulator there are no electrons in the upper conduction bands. Multichannel Analyser. Hewlett-Packard 0-200 V. Incident γ E Conduction band Band gap (6 eV for NaI) Thallium levels Valence band ejected electron leaves a hole The ejected electron leaves a hole in the valence band .and this causes the hole to move through the crystal.E E E 2 1 r crystalline (small r) gaseous (large r) The electrons with which photons produced by the 137 Cs and 60 Co gamma-ray sources used in this experiment will collide belong to the N aI crystal doped with Thallium in the scintillation counter. In the drop from the conduction band intothe excited thallium electronic level. Neighboring atoms try to share electrons in attempts to fill in the hole. A gamma ray photon can collide with an electron in the valence band of the NaI crystal. a photon of a 131 . giving it enough energy to be ejected into the band of conduction energy levels or even higher. it can ionize the Thallium atom and the previously ejected electron can fall into one of the excited states of the thallium atom. If the hole encounters a Thallium atom impurity. This electronic pulseis delivered to an amplifier. The photoelectric effect is used in an important bit of laboratory apparatus called the photomultiplier. performing an otherwise forbidden transition and emitting light. This electron is accelerated by around 150 V by a high voltage 132 . The primary electron rejoins the valence band by falling into these intermediate levels. NaI Photocathode Dynodes The phototube then increases the number of liberated electrons 106 fold by producing secondary electron emissions with a series of dynodes. The light produced is reflected by alumina or M gO reflective layers onto a photocathode that emits electrons by the photoelectric effect. and so the number of pulses displayed versus amplitude (channel or energy) corresponds closely to the energy spectrum of the primary radiation. To a very good approximation the pulse created is directly proportional to the energy dissipated in the scintillator. they look like the following figure. This device has a window coated with a photosensitive material that ejects an electron when a photon of sufficient energy hits the window. The electron ejected from the valence band by the γ-ray is called the primary electron. Schematically.few eV energy is emitted. and doping with Thallium provides intermediate energy levels in the forbidden energy region between valence and conduction bands. that starts the entire scintillation process. The figure above shows the spectrum recorded by a NaI(Tl) crystal when exposed to 0. a typical gain factor for PM tubes operating at about 2000 V . Gamma ray Photo-cathode Dynode Distributor (high voltage to dynodes) High voltage input Signal output The probability that the origin photon entering the tube actually ejects an electron from the photocathode is called the quantum efficiency. higher voltage dynode. For light in the visible range. The accelerated electrodes cascade deeper into the nest of dynodes. This is the first stage of a multi-stage collection of dynodes. Eventually the single electron has become 106 electrons. liberating more electrons.6616 MeV γ-rays.25. Ideally we should get a continuous signal extending from zero energy to 133 . This many electrons produces a signal big enough for electronics to process and record. a good PM tube has a quantum efficiency of around 0. which it strikes. where it liberates more electrons.electrode. The accelerated electron proceeds to the next. 11010 counts and ch2 = 947. due to secondary collisions and e+ e− pair creations.4 Stage 1. For this we use the 60 Co source which emits γ−rays of energies E1 = 1173. E is the energy and a and b are constants to be determined by calibrating the device as illustrated in a previous section. In practice the Compton continuum is smeared out.35σ 134 . the half-height channels are deviated from the photopeak channels by √ ch = chp ± 2 ln 2σ = chp ± 2. 13. and the count number versus channel will look like the figure below. ch12 = 873 ch21 = 907. the standard deviations given by (ch−chp )2 1 A = Ae− 2σ2 2 or. ch2 by looking for the channels on either side of the peak that hold half of these numbers of counts.002KeV To calibrate the MCA.470 ± 0. ch22 = 985 Assuming that each peak is a normal curve. The Multichannel Analyser displays 1023 channels and plots the number of gamma rays detected per channel as a graph. The Compton Edge or the channel at which electrons are ejected with Kmax = Ee by the γ-ray. This is the energy that we are looking for in this lab. the energy E1 photopeak occured at ch1 = 837. and a sharp peak at Eγ for a single gamma ray energy incident on the scintillator. we will count the cobalt source for about 4 hours. 8408 counts We compute the standard deviations of the values of ch1 . The relation between the channel at which the photon is detected and the corresponding energy deposited by the event in the scintillator is linear ch = a + bE in which ch is the channel. ch11 = 797. the following was found.002KeV and E2 = 1332. is halfway up to the Compton plateau.210 ± 0. calibration When the experiment was performed in 1995. as is the photopeak.Kmax = Ee . 470) and determine that the calibration constants are a = 26.and so σch1 = ch12 − ch11 = 32. b = 0.35 and We then set up the linear equations ch2 = 947 ± 33. the electron can acquire energies of several MeV. ch1 = 837 ± 32.54 keV agreeing well with the 661. 000counts ch31 = 453.64 keV accepted value.34.64keV was found with ch3 = 476. By examining the energy of the Compton edge and the photopeak from the scintillator.318 keV . ch32 = 496 and so and a Compton edge at corresponding to an energy ch3 = 476 ± 18. The photopeak corresponds to a γ-ray of energy Eγ = 650. 14 Energy versus Momentum for Relativistic Electrons When an electron in a scintillators valence band is struck by a γ-ray from a radioactive source. 60.67. photopeak and edge measurements The 137 Cs sample was then set up and the following photopeak spectrum was observed after counting for 30 minutes.19 837 = a + b(1173.210).3 che = 345 ± 10 Eedge = 461keV which is quite close to the accepted value of 477. one can easily verify the form of the relativistic dispersion law E 2 = p 2 c2 + m 2 c4 135 . making them highly relativistic. 947 = a + b(1332.6907 13.34 2.5 Stage II. The main photopeak corresponding to energy E3 = 661. or Ee + mc2 . 54 M n.178 MeV From this data the channel (ch) versus energy (E) calibration constants for the multichannel analyser are determined ch = a + bE We found in the last expriment that the electron kinetic energy at the Compton edge is 2Eγ2 Ee = mc2 + 2Eγ and at the Compton edge. Since the photon is backscattered we find that Eγ − (−Eγ 0 ) Pe = c Eliminating Eγ 0 we find that after the collision the electron momentum is Pe = 2Eγ − Ee c where Ee is the Compton edge energy and Eγ is the photopeak energy. 137 Cs.511 MeV 0. which presumably satisfies (Ee + mc2 )2 = Pe2 c2 + m2 c4 = Ee2 + 2Ee mc2 + m2 c4 136 .835 MeV 1. the backscattered photon has energy Eγ 0 = mc2 Eγ mc2 + 2Eγ Adding we find that Ee + E γ 0 = E γ equaling the energy of the pre-collision photon. with additional 22 N a.1 Apparatus The same as the Compton edge experiment.622 MeV 0. The total energy of the electron is kinetic plus rest. The energies Eγ of the γ rays emitted by each source can be found below. 14.2 Procedure The Compton edge and photopeak energies of all of the samples are measured.14. and 60 Co γ-ray sources. The actual photopeaks should occur at Isotope 22 Na 137 Cs 54 Mn 60 Co Eγ 0. a1 .03806 1.002837 + 0.640 0.or Ee2 + 2mc2 Ee = Pe2 c2 .88236 We now run a nonlinear least squares fit of this data to try to fit the best curve to it.7 212. we will measure Ee and Eγ and plot 2m versus e Ee . a2 . Pe2 Ee2 = Ee + 2m 2mc2 2 Pe In order to determine the true dispersion law. and a2 = 0.000035 M eV )c + (0.493 M eV 2(1. We obtain the estimate 1 me c 2 = M eV = 0.3 Data and analysis The channels and corresponding energies for photopeaks and Compton edges for all four sources are in the table below.7 137. Isotope 22 Na 137 Cs 54 Mn 60 Co cγ Eγ M eV 107. obtaining Pe2 = 0. assuming that Pe2 = a0 + a1 Ee + a2 Ee2 2me and run a nonlinear least squares fit to determine a0 .835 260.341 0. Classically we should find that a0 = 0.969 Pe2 2me M eV 0.969 We are using the MCA calibration constants E = (0. a1 = 1.621535 1.5 0.0 0.5 93.341 0.640 0. 137 .00438 ± 0. 14.6%. however what we will see is relativistic departures from this dispersion law.511 143. I used the program polyfit to perform the regression.5 0.9 Ee M eV 0.015000Ee2 2me in extremely good agreement with a supposition of a relativistic dispersion relation.957349 Ee + 1.453778 0.0 1.622 183.015) which is in error by only 3.178 ce 69.0366 M eV ) found by matching the photopeaks to their known energies.447 0.447 0. We now compute the ”classical energy” Isotope 22 Na 137 Cs 54 Mn 60 Co Ee M eV 0. 15 Michelson-Interferometer and the Index of Refraction of Air Apparatus 1. He-Ne Laser. Beck Interferometer M300/6407. 4. Vacuum pump. 138 . Chart recorder. The Beck interferometer used in the Michelson mode looks like the following. 5. 3. Phototransistor. 2. air cell. however this apparatus belongs to the chemistry department and is used in the physical chemistry lab course. The department has it’s own Michelson interferometer. as in A below. The eye is aligned with the viewing axis about 10 inches from the instrument. and the control should be adjusted so that the moveable image superimposes on the stationary one B.M 2 Air Cell Source To Vacuum pump M3 Compensator M 1 Operation of the interferometer is simple. 139 . There will be three reflected images of a pointer in the beam divider head. Adjust the tilt controls on M1 to see which image can be moved. A photocell and chart recorder are used to count the number of wavelength shifts the insrutment cycles through as the cell is evacuated. Center this collection of fringes. order n − 1. The instrument is now ready and the path length is zero. If the mirrors are parallel but the path lengths are unequal. the fringes will be circular with a center within the field of view. n − 2 and so forth of increasing radii.A C B D When exact vertical on horizontal coincidence of the images is achieved the interference fringes will become visible. Turn the path length control dial in the upper right corner of the top figure in whatever direction causes the fringes to move inward. Now adjust the tilt controls to produce about 10 nearly vertical fringes as in D.8173 × 10−9 m 140 . order n in the middle. Zero path difference can be calibrated with a Mercury vapor lamp.10 m. interference fringes will even be visible with white light from a tungsten filament. Procedure The He-Ne laser is split and recombined after passing through a pressurized cell. Sample data The length of the air cell used is 0. the laser has a wavelength of λ = 632. the fringes contract to a central point. the fringes will be sections of circles surrounding a center displaced from the field of view. and then slowly repressurized. Now switch to a tungsten lamp and continue to turn the path length control very slowly until a group of bright fringes with a distinctive central dark band in the middle appears in the field. As the path difference becomes zero. For unequal path lengths and nonparallel mirrors. in the direction of decreasing radius. If the path lengths are equal to within ±5 fringes. and adjust the path length control until the fringes are straight. 141 . 047 which agrees very well with the accepted value of 1. Four separate trials revealed 83 fringes were shifted through as the cell was refilled with air.8173 × 10−9 m waves in the cell in vacuum. 047 = N λ 632. 047 + 83 nair = = 1.We find that there will fit 0.0002626 316. and so 316. the air temperature of the experiment.2m 2L = = 316.0002684 at 70o F . and so 2L = N λv = (N + N 0 )c Nc = ((N + N 0 )λair ) = ν nair ν where N 0 is the number of wavelengths the interferometer cycles through as the cell is refilled. σ0 The part of a properly quantized photon vector potential responsible for creating a planewave photon will then look like A= Z d3 k 0 X σ 0 =1.m create an electron with wavefunction ψn.σ create a photon with wavevector k and (transverse) polarization vector v k.2 · k = 0.`.`.m . Let a†n. We can treat this problem using perturbation theory and the Golden Rule developed earlier.`.`0 . Both of these field operatorscan be Fourier expanded in terms of solutions to the appropriate Schroedinger or Maxwell equation. `0 . The problem of energy emission by the atom is most appropriately addressed through quantum field theory. and we will do so in a very schematic manner.`.m0 ] = δn. can be gotten from the Bohr formula hν = ∆E = − Z 2 mc2 α2 1 1 ( 2 − 2) 2 Ni Nf however not all energy level transitions that one might naively assume occur do not in fact ever happen.m (r.`0 δm. so that [an. if we model the interaction between the photon and the atom with HI = Z A · J d3 x h ¯ (Ψ∗ ∇ − (∇Ψ∗ )Ψ) is a “field operator” for the electronic current and in which J = −q 2im A is the vector potential of the photon field. mi → |n0 . m0 i ⊗ |k.σ0 = δσ. `. φ) from the vacuum.m0 and let b†k. a†n0 .n0 δ`. Emission of light is a creation process. θ. vk.2 Nk0 .σ · vk.1 · k = 0. and an.m be the corresponding annihilation operator.16 Angular distribution of emitted radiation The energies. vk. of light emitted by the Hydrogen atom when it undergoes a transition from one energy level to another. Recall that in empty space the independent polarization vectors must be perpendicular to the wavevector vk. and therefore frequencies. a theory developed to study processes in which particle number or number of quanta is not preserved. |n.σ0 eik ·r b†k0 . σi in which the rightmost ket is that describing a photon of wavevector k and polarization state σ. because the transition violates certain conservation laws.σ0 0 142 .σ . We not only have an energy balance to preserve Ei = Ef + hν but angular momentum must be preserved as well as linear momentum. because we are in fact getting far ahead of ourselves right now as far as subject matter is concerned.σ0 vk0 . `f .m00 |ni . θ.`f .σ0 ·(ψn∗ 0 . mi i = |0iδni . We can now apply Fermi’s Golden Rule which states that the probability of transition from initial state |ii to final state |f i is proportional to the modulus of the matrix element hf |HI |ii. k. m|n0 . `i . φ)an0 . φ)a†n0 . θ. `f . we only need to reatin the one term in four that has a nonvanishing matrix element between the one electron in and out states.`0 . σ| bk0 . `i .m00 Z d3 k 0 X σ 0 =1.m0 ) X n0 .mi − ∇ψn∗ f .`00 . We find then that allowed transitions resulting in the emission of light will have nonzero values of Z d3 x hnf .`00 . mi i When we exand J in terms of annihilation/creation operators. and wavefunctions are matrix elements of field operators in the basis of such particle number states.`0 .n00 δ`i .`00 δmi .m0 (r.`0 .`. `i .m0 δk.`00 .σ0 vk0 .m0 (r.m0 When acting on the vacuum. m|Ψ|0i = X n0 .σ eik·r There will be a similar Fourier expansion for the electron field Ψ= (ψn∗ 0 .ell00 .σ · (ψn∗ f . φ) We have performed a rather subtle conversion to the Bargmann-Fock space in which states are labeled by the number of quanta of each type occupying the universe.2 0 Nk0 .m0 ∇ψn00 . this will create a Hydrogenic bound state particle with wavefunction hn. `0 .m00 and we finally obtain the matrix element ℘∝| Z d3 x e−ik·r (− q¯ h )vk.mi )|2 2mi which is basically just the Fourier transform of the electronic current matrix element between the initial and final electron states.m0 (r. mf .`0 δmf . φ)hn.`i .m (r.`i .σ0 and an00 .`f .m0 ψn00 . σ|k 0 .σ0 a†n0 .2 0 Nk0 .`0 .mf ∇ψni . `f . we find then that the expression above reduces to Z 3 dx X X n0 .mf ψni . `.`0 . σ| bk0 .`0 .m00 |ni .`0 . m0 i = ψn.`0 .`0 . mf .σ0 e−ik ·r (− q¯ h )vk0 .m0 n00 . k.σ0 eik ·r hk. mf .k0 δσ.`0 .m0 ψn0 .m0 + ψn0 .m0 = h0|δnf .`00 . σ|A|0i = d3 k 0 X σ 0 =1. `. mi i which we now evaluate using hnf . σ|A · J|ni . 143 . σ 0 i = vk.m00 ) 2mi ·hnf .m0 an00 .which when acting on the vacuum creates a photon with wavefunction Z hk.m00 −∇ψn∗ 0 . k. θ.`0 . θ.`0 .n0 δ`f .σ0 a†n0 . it is the angular wavefunctions that really determine the selection rules since s 8π <e Y1. yfy |∞ −∞ .`f .`i .mi ) |2 which is proportional to the matrix element of the dipole operator qr between the initial and final states. zfz |∞ −∞ ) − Z 3 (fx . fz ) d x = − Z f d3 x Applying this now to our current operator shows that Z (ψi ∇ψf∗ − ψf∗ ∇ψi ) d3 x = 2m(Ef − Ei ) h ¯2 Z ψf∗ r ψi d3 x the transition probability is then.1 (θ. Transitions resulting in the emission of light between two states ψi and ψf are said to be forbidden if the matrix element of the electric dipole operator between these two states is zero. For two different solutions to the Schroedinger equation (− h ¯2 2 ∇ + V (r))ψf∗ = Ef ψf∗ . Z 3 r∇·f d x = Z (x. φ) x = r cos φ sin θ = −r 3 144 . Whether or not this matrix element is zero has little or nothing to do with the radial part of the wavefunction. fy . y. e−ik·r ≈ 1 within the volume over which the bound state wavefunction probability density is appreciable.σ · (− 2 2mi h ¯ Z d3 x (ψn∗ f .mf r ψni . in the dipole approximation 2m(Ef − Ei ) q¯ h ℘∝| )vk. z) ( ∂fx ∂fy ∂fz 3 + + )d x ∂x ∂y ∂z which can be integrated by parts to give = (xfx |∞ −∞ .The Schroedinger equation itself allows us to evaluate this matrix element in the dipole approximation. 2m (− h ¯2 2 ∇ + V (r))ψi = Ei ψi 2m we multiply the first by ψi and the second by ψf∗ and subtract to obtain h ¯2 ∇ · (ψi ∇ψf∗ − ψf∗ ∇ψi ) = (Ef − Ei )ψf∗ ψi − 2m Multiply both sides and integrate − h ¯2 2m Z d3 x r ∇ · (ψi ∇ψf∗ − ψf∗ ∇ψi ) = (Ef − Ei ) Z d3 x rψf∗ ψi However for any vector f that vanishes sufficiently strongly at infinity. y = r sin θ sin φ = −r and s 8π =m Y1. mi i + B 00 |nβ . cos θ sin φ. σ = 1. which we will take to be vk. φ) 3 and we can use the Clebsch-Gordon recursion relations deried much earlier in our study of angular momentum to deduce that z = r cos θ = r x|ni . mi ± 1i + B|nβ .2 = (− sin φ. `i . mi ± 1i y|ni . `i + 1. mi ± 1i + B 0 |nβ .σ . 0) which satisfy vk. `i + 1. φ) 3 s 4π Y1.0 145 . − sin θ) and vk. sin θ sin φ. mi i = A00 |nα .`i .σ · 2 2mi h ¯ Z d3 x (ψn∗ f . φ) = Ri (r) Y0. `i + 1.mf r ψni .σ · k = 0. You may have noticed that Fermi’s Golden Rule even gives us some indication as to the polarization state of the light emitted in a given transition. For light with k-vector k = kn. mi i = A0 |nα .1 = (cos θ cos φ.1 (θ. `i − 1. m f = mi These are known as the electric dipole selection rules and they govern what we see in the spectrum of atomic transitions. and we can determine the probability that a photon emitted in a certain direction n = (sin θ cos φ. `i .`f . `i − 1.mi ) |2 we can determine the most probable direction of emission for an oriented atom or nucleus undergoing a transition. mi ± 1i and z|ni . `i − 1. there will be two independent polarizations vk. `i .1 Polarization and angular distributions Using the formula ℘∝| 2m(Ef − Ei ) q¯ h (− )vk. cos θ) will have a given polarization state. mi i = A|nα . If our atom or nucleus begins in a state with wavefunction ψi (r.0 (θ. 2. 16. mi i These tell us that the final state of the atom must differ from the initial by `f = ` i ± 1 and mf = mi ± 1. θ. cos φ. 1 + δm.−1 )+(cos θ sin φ)( √ )(δm.mi ) = Rif ( − sin φ √ (δm.mf ∞ 0 ∞ 0 1 d z ψni .−1 ) 6 6 Rif = √ ( − ie−iφ δm.1 + δm.mi ) = Rif ((cos θ cos φ) √ (δm.`i .m with m = 0.1 · d3 x (ψn∗ f .−1 ) 6 6 Z 1 − sin θ √ δm. ±1.−1 ) Ri∗ (r) rRf (r)r 2 dr(δm.1 − δm.mf x ψni .`f .0 ) 3 √ Rif = √ ( cos θe−iφ δm.`f .`f .−1 ) 6 If we place a detector to receive light emitted in the n direction.−1 ) Z ∞ 0 Ri∗ (r) rRf (r)r 2 dr δm.0 Ri∗ (r) rRf (r)r 2 dr We can now construct the probability that an oriented atom will emit light in the k = kn direction with polarization states 1 or 2.−1 ).`i .−1 ).1 −δm.`f .1 − δm.mi ) = √ 6 Z d 3 x (ψn∗ f . First we compute 1 −i vk. regardless of polarization.mi ) = √ 6 Z −i y ψni .1 − Y1.mf r ψni .0 ) 6 and for the other polarization vk.−1 − 2 sin θδm. If we use s 2π x=r (Y1.`i . θ.1 + Y1.1 +δm.0 3 then our matrix elements will be Z Z and 1 d3 x (ψn∗ f .2 · Z 1 −i d3 x (ψn∗ f .`i .2 ℘0→0 = | √ q¯ h 2 Rif 2 2m(Ef − Ei ) √ 2 sin θ|2 | | − (− )| | 2mi h ¯2 6 146 .we have seen from the dipole selection rules that the final state after photon emission must be ψf (r.mi ) = √ 3 and we will use the abbreviation Z 3 x (ψn∗ f .mf Rif = Z ∞ 0 Ri∗ (r) rRf (r)r 2 dr(δm.−1 ) + cos φ( √ )(δm.1 + cos θeiφ δm. the probability that the transition into the m = 0 state will result in light of any polarization arriving at the detector is X σ=1. φ) = Rf (r) Y1.mf r ψni . 3 z=r s 4π Y1.`f . 3 s 2π y = −ir (Y1.`i .1 + ieiφ δm. m).0) The intermediate state has three possible orientations with respect to an arbitrary quantization axis. a (l. we would of course have different matrix elements.2 ℘0→+1 = | 2m(Ef − Ei ) q¯ h 2 Rif 2 )| | √ | (| cos θeiφ |2 + | − ieiφ |2 ) (− 2 2mi h ¯ 6 2m(Ef − Ei ) q¯ h 2 Rif 2 1 + cos2 θ )| | √ | ( ) (− 2mi 2 h ¯2 6 If the transitions were from levels or into levels with ` 6= 0.m)=(0. which must be a ∆m = −1 → 0 or ∆m = 1 → 0. and could end up with more complicated angular dependencies on our emission probabilities.and the probability that the transition into the m = −1 state will result in light of any polarization arriving at the detector is X σ=1. If we set up a detector as illustrated below. =| The rather good text Experiments in Modern Physics by Melissinos gives the following concrete example.2 ℘0→−1 = | q¯ h 2 Rif 2 2m(Ef − Ei ) (− )| | √ | (| cos θeiφ |2 + |ieiφ |2 ) 2 2mi h ¯ 6 2m(Ef − Ei ) q¯ h 2 Rif 2 1 + cos2 θ )| | √ | ( ) (− 2mi 2 h ¯2 6 and for the m = 1 we get the same result =| X σ=1. 1. the direction between the sample of emitters and the detector. The first γ for a → b will be emitted isotropically.m)=(0. about which we have no knowledge. Now this nucleus performs its second decay. m=0.+1 bc transition c (l.0) ab transition b (l.-1. and so has an anisotropic distribution 1 + cos2 θ ℘(θ) ∝ ( ) 2 or ℘(θ) 1 + cos2 θ ) = ( ℘( π2 ) 2 147 . establishes a quantization axis which we take to be the z-axis. Consider a nucleus or atom that decays from one quantum state to another in two stages as shown below. When this detector gets a γ of energy Ea − Eb .m)=(1. receiving the first photon. it came from a nucleus left with its angular momentum vector having z-component m = ±1 with respect to this axis (since sin2 0 = 0). for which we can use our computations above. meaning related.172 M eV γ and then a 1. Photomultiplier 1 Gamma source z θ n Photomultiplier 2 The decays that produce these γ rays are between high-angular momentum levels. The direction of the first emitted γ will be used as the z-axis for a study of the angular distribution of the second γ. we will use nuclei that emit two gamma rays. and apply a strong magnetic field. How can we orient an atom or nucleus? We could freeze it down to quench rotations and vibrations. and when the second arrives at the moveable counter. Cobalt-60 decays to Nickel-60. 10 − 15 degrees apart. coming from the same source. and so the circuit must be tuned to count coincident γ’s with this time delay between arrivals at the respective PM tubes. The delay for 60 Co is about 1 × 10−12 s.333 M eV γ. which in turn emits first a 1. that they cme from the same nucleus? The signals from both PM tubes are fed into a coincidence circuit.2 Angular correlations of γ emissions What we have said above applies to an oriented emitter. In the nuclear γγ angular correlation experiment. The odds are pretty good that simultaneously arriving γs of the right energy are correlated. How do we know that when one γ of the first emission energy arrives at the fixed counter (photomultiplier 1). with a delay of only 1 × 10−12 s between these emissions.16. 148 . which only registers simultaneous γ arrivals. and a detailed computation of the angular distribution of the second photon would yield a theoretical angular distribution 1 1 ℘(θ) = 1 + cos2 θ + cos4 θ 8 24 This could be measured by moving the angular placement of a second photomultiplier tube from one angular position θ to another. and counting at each location for equal times. 000000 180. The PM tubes each subtend an angle dθ = 8. and it gives us the following data. Two photomultiplier tubes.000000 PM-1 integration 13471 13315 13173 13105 13254 13420 13286 13146 13465 13249 13454 13451 13417 13217 13422 13269 13260 13086 13221 13555 PM-2 integration 13555 13433 13208 13378 13423 13204 13291 13413 13204 13590 13395 13314 13394 13450 13127 13358 13247 13340 13350 13570 C(θ) coincident counts 14 10 9 11 14 16 20 26 26 36 45 73 90 88 121 199 339 604 1386 2967 The simulation confirms the hypothesis that simultaneously emitted γ rays are emitted from the same nucleus. at at 180 degrees apart. We will need to know the distance of the PM tubes from the source. It uses the accept-reject algorithm to produce a random number sharply peaked (Lorentzian peak) about zero. in order to compute the angular width of the tubes as subtended from the source location.16. tube 2 is on a traveller that can be stopped at 10 degree position intervals. the angle θ is the position of the moveable PM.000000 153. Apparatus would consist of 1. 2.000000 108. The program used for the simulation is listed below. and we integrate then from 150 to 201 keV for each PM.000000 144. equidistant from a 60 Co source. This means that it is integrating the signal between one full peak standard deviation on either side of the peak.000000 36.000000 72. consistent with conservation of momentum for pair annihilation.000000 81. Angle θ 9. Tube 1 is fixed. Coincidence counter.000000 27.000000 45. The counting time is set at 100 s for each angular position.3 A simulated experiment I have not performed this experiment myself.000000 126.000000 135.000000 117. The coincidence circuit will count all events within a certain channel number of the photopeak.000000 54. which is used to 149 .000000 90.000000 171.000000 99. but we can easily simulate it to see precisely what the data should look like. The peak occurs at a channel corresponding to energy Eγ = 178 keV .000000 18.000000 162.000000 63.0 degrees as measured from the source. h> #define PI 3.h> #include<math.theta)==1) c=c+1.create a direction for the second photon once the first photon random direction has been computed.0+d+lorentz().c1.0. 150 . /* gamma-gamma correlation simulation */ #include<stdio.0 /* angular width in degrees */ long seed. srand(seed). c2=0. dp=180. /* d.m++){ theta=(double)m*del.fabs(d-dp)).m. /* detect if PM is triggered */ int trigger2(double angle. /* angular increments for moveable PM */ /* counters set to 0 */ for(m=1.} /*printf("%f\t%f\t%f\n".dp.n<300000.n++){ d=360.c2. while(dp>360. double d. del=9. /* get a lorentzian random between 0. 360 peaked at 180 with respect to th*/ main(){ seed=1342.0*(double)rand()/(double)RAND_MAX. double pos).h> #include<stdlib.1415926 #define dtheta 8.*/ if(trigger1(d)==1 && trigger2(dp. c1=0.0){dp=dp-360. int n. d. c=0. int trigger1(double angle). for(n=0.dp.del.m<=20. c is coincidence tally */ double lorentz().0. dp are two correlated directions */ /* dtheta is angular width of detectors */ /* theta is moveable detector position */ long c. /* ci is counter i tally.theta. c1. } printf("%f & %d & %d & %d\\\\\n".num2. } double y11cor(){ double num.c). } double lorentz(){ double num. 180 */ /* peaked sharply around 0 */ start: 151 .0*(2. }/*end loop over angles */ } int trigger1(double angle){ if(angle >= 0.test.if(trigger1(dp)==1 && trigger2(d.0) return(1). } int trigger2(double angle. theta.num2.theta)==1) c=c+1.theta)==1) c2=c2+1. if(trigger1(d)==1 || trigger1(dp)==1) c1=c1+1.0+num*num). double pos){ if(angle >= pos-dtheta/2. if(trigger2(d. num2=(double)rand()/(double)RAND_MAX. else return(0). 180 */ /* peaked sharply around 0 */ start: num=180. test=100.0/(100. else goto start. else return(0).0 && angle <= dtheta) return(1). /* get a random angular change between -180.c2. if(num2<=test) return(num). /* get a random angular change between -180.0).test.0*(double)rand()/(double)RAND_MAX-1.theta)==1 || trigger2(dp.0 && angle <= pos+dtheta/2. num=180.0*(2.0*(double)rand()/(double)RAND_MAX-1.0); test=0.5*(1.0+cos(num)*cos(num)); num2=(double)rand()/(double)RAND_MAX; if(num2<=test) return(num); else goto start; } A second unused routine is provided to generate random angles in degrees with PDF f (θ) = 1 (1 + cos2 θ). 2 We now work up our data by producing a polar plot of coincidence versus angle, which translates into coincidence being the radial coordinate. The therefore plot the points (x, y) = (℘(θ) cos(θ), ℘(θ) sin θ) ∝ (C(θ) cos(θ), C(θ) sin θ) Which should be a spike directed to the left from the origin. 22 Na γ γ correlations 1.0 C(θ) sin(θ) 0.8 0.6 0.4 0.2 0.0 −1.0 −0.8 −0.6 −0.4 C(θ) cos(θ) −0.2 0.0 As a good exercise you are urged to modify the simulation to produce data consistent with the angular correlations of emissions from the Cobalt source. The polar plot should look like the following. 152 60 Co γ γ correlations C(θ) sin(θ) 2.0 1.5 1.0 0.5 0.0 −1.0 −0.5 0.0 0.5 C(θ) cos(θ) 1.0 and those for the example used in the computations above, with 1 C(θ) = (1 + cos2 θ) 2 would look like the following; dipolar γ γ correlations 2.0 C(θ) sin(θ) 1.5 1.0 0.5 0.0 −1.0 −0.5 0.0 C(θ) cos(θ) 153 0.5 1.0 17 A simulated experiment; Geiger-Marsden experiment In nuclear and high energy physics, which deal with probing the structure of very small objects, the tool of exploration is scattering. We see the shapes of objects by observing either light directly emitted by luminous objects, or by seeing light scattered off of the object from some other source. In order to see into very small objects, light or other particles of very small de’ Broglie wavelength must be used. Fairly long ago the technology to produce beams of particles such as alpha-particles (Helium nuclei) was developed and used to probe into the interior of atoms, in order to determine how the atom is constructed. The definitive experiments were conducted by Rutherford and Marsden. Consider a beam of hard sphere particles colliding with a collection of targets, with a fixed target density per unit volume. If a projectile can hit a certain area A(θ) around any target in the entire length L of the target sample, it will be scattered by a scattering angle between p θ and θ + dθ. If we have an incoming flow of projectiles at rate Rp = dN distributed unidt formly and flowing through an area A, we have a particle beam A A N target L of flux F = RAp . The rate then with which projectiles scatter by angle θ to θ+dθ is determined by how many will hit the necessary area Ntarget · A(θ). This will be dN (θ) = Rscatt (θ) = F · Ntarget · A(θ) dt 154 N A target A The area A(θ) is called the differential cross section, and is an annular region within the confines of a certain impact parameter b and b + db, so that A(θ) = 2πb db, which we have projected onto the targets in the two figures above. Consider classical scattering of particles of mass m, initial speed v0 and impact parameter b by a static central force field. If all particles with impact parameter in the range b → b + db are scattered into solid angle φ θ b m dθ Ω → Ω + dΩ. What is the meaning of the so called scattering cross section? It represents the size of the target that an incoming projectile must hit in order to be scattered through an angle between θ and θ + dθ. For a given target then if a projectile hits the 155 ring of area 2πb db in the figure, it will scatter through such an angle. Suppose that an incoming particle beam has a flux of F particles per second per unit area. The rate with which projectiles scatter into θ → θ + dθ is d2 N = F · 2πb db dt We translate this from a function of b to a function of θ with db d2 N = F · 2πb dθ dt dθ and now find the number of particles per unit time scattered into the cone of outer angle θ + dθ and inner angle θ, which subtends a solid angle dΩ = 2π sin θ dθ; db 1 b db d2 N = F ·b (2π sin θ dθ) = F · dΩ dt dθ sin θ sin θ dθ and finally we define that quantity loved by particle physicists; the differential cross section 1 d2 N b db dσ(θ) = = dΩ F dt dΩ sin θ dθ The trajectory of the particle can be gotten from the energy expression m 2 (r˙ + r 2 φ˙ 2 ) + V (r) 2 but we can eliminate t in favor of φ using the angular momentum equation E= ` = mr 2 dφ dt or by inverting d ` d = dt mr 2 dφ We find that E= `2 `2 dr 2 ( ) + + V (R) 2mr 4 dφ 2mr 2 or dφ = Now set u = 1 r `dr mr 2 and φ1 − φ 0 = q Z 2 (E m − V (r) − u1 u0 q 2mE `2 156 −du − 2mV `2 `2 ) mr 2 − u2 If we let the initial and final values of r be ∞ and the point of closest approach be r0 , then we find that φ1 = π, φ0 = θ + ψ and π = 2ψ + θ and Z θ =π−2 bdu umax q 0 with the angular momentum given by 1− V (u) E − b 2 u2 √ ` = mbv0 = b 2mE We now specialize to the case of the coulombic potential V (u) = ku and compute the point of closest approach 1 r0 = umax The radial speed is zero at this point and so r˙ = 0 and E = kumax + l2 u2max 2m √ k 2 + 4b2 E 2 2b2 E We can now compute the integral for the scattering angle and obtain umax = −k + θ = π − 2 sin−1 (1) − 2 sin−1 ( √ k2 k ) + 4b2 E 2 which can be solved for b in terms of θ θ k cot( ) 2E 2 which in turn can be inserted into the cross section formula b= k2 θ dσ(θ) = csc4 ( ) 2 dΩ 16E a An alternative approach that does not involve integral transformation is to use the orbital equation to find the angle between the asymptotes of the hyperbolic path taken by the projectile. For Coulombic scattering from a nucleus, V (r) = k r with k > 0, d ` k 1 E = m(( r)2 + r 2 ( 2 )2 ) + 2 dt mr r the point of closest approach is, s mk mk 2mE = = − 2 + ( 2 )2 + 2 rmin umax ` ` ` We know that paths will be hyperbolas, such as in the figure 1 1 157 y φ 0 x u max Θ 1 = A − B cos θ r with asymptotes at θ = ±φ0 so use conservation of angular momentum u= d ` d = dt mr 2 dθ to write 1 d d k 1 ` dr `2 k E = m(( r)2 + (r θ)2 ) + = m(( 2 )2 + 2 2 ) + 2 dt dt r 2 mr dθ mr r 1 again use u = r to get ` du 2 `2 1 ) + 2 u2 ) + ku E = m(( 2 m dθ m Insert our hyperbola solution to get E= `2 `2 2 B (1 − cos2 θ) + (A2 − 2AB cos θ + B 2 cos2 θ) + k(A − B cos θ) 2m 2m The coefficient of cos θ must be zero; 0 = −kB − or A=− and so s B=− ( `2 AB m mk `2 mk 2 2mE ) + 2 `2 ` 158 and we have our complete path. To get the asymptotes we note that at a point where r → ∞, u → 0 and so at θ = ±φ 0 = A − B cos φ0 and so cos φ0 = or A 1 =q 2 B 1 + 2E` mk 2 2E`2 mk 2 but the scattering angle is related to the asymtote angle by tan2 φ0 = 2φ0 + π = θ so `= s mk 2 tan φ0 = 2E s θ mk 2 cot 2E 2 Remember that 1 E = m v02 , 2 where b is the impact parameter. ` = m v0 b The cross section derived above applies to one target. Suppose a beam of projectiles is shot at a target that consists of many nuclei, such as a gold sheet? The number of gold nuclei per unit volume is ρ, and so the number of targets per unit volume is ntarget = ρNA M where M is the atomic mass of gold, and NA is Avagadro’s number. If the target is gold foil of cross-sectional area A and thickness L, we have Ntarget = ρNA AL M gold nuclei, which we multiply our cross section computed earlier by, to arrive at d2 N db 1 = F · Ntarget · b (2π sin θ dθ) dt dθ sin θ This quantity can be measured experimentally; if we bombard the target for time T we will have accumulated db 1 dN = F · Ntarget · T · b (2π sin θ dθ) dθ sin θ scattered projectiles in the detector positioned to count projectiles scattered into the cone of solid angle 2π sin θ dθ centered on angle θ. Rutherford, Marsden and Geiger bombarded thin gold foil with alpha particles. They were watching for substantial back scattering at angles close to 180 degrees, since the plum pudding model lacked a hard, localized positive scattering center there should be practically 159 The entire inner wall is lined with photographic film. was very potent by modern standards. in which a fine powder of the crystal is bombarded with X-rays. Rutherford analyzed the data and confirmed that it supported the hypothesis that the atom possesses a concentrated and very small nucleus.no backscattered alphas. The scattered X-rays expose the film. such as that used by Rutherford. The details of this very thorough and brilliantly executed experiment are quite remarkable considering the era in which it was performed. and was visually monitored with a microscope. Bombardment of a thin film of gold. a butt-kicking 0. rather than the 1 in e(90) predicted by the plum pudding model. the angular displacements of diffraction maxima can be found. and X-rays enter through a port. This is enough to emit several billion α particles per second in the decay process 222 Rn →218 P o + α The detector used was a micrometer-mounted zinc-sulfide screen that would briefly flash when an α struck it. Rutherford himself was surprised and compared it to firing a 15 inch naval artillary shell at a sheet of tissue and having the shell bounce back at you. a sample of purified radium. To explain this required all of the positive charge in the atom to be localized at a single very massive point with the electrons orbiting in a cloud in sufficient numbers to make the atom neutral. The classical scattering formula for a pure Coulombic nuclear potential gives very good agreement with the experimental data. The screen could be moved from one angular position to another. with α particles (42 He nuclei) requires an apparatus very similar to that used in the Bragg scattering experiment.000 alphas scatter through an2 gles exceeding 90 degrees. essentially an X-ray version of Bragg scattering. Geiger and Marsden. scattering angle. 160 . The experimenters studied the scattering rate as a function of incoming α particle energy.1 Curie. Chemists often determine the lattice spacings of crystalline samples using a powder diffraction apparatus. The α source used. The sample is held at the center of a flat cylindrical disk container with a removable top plate. and several other factors. and by careful measurements of the positions of exposed areas. They found instead that 1 in 10. which uses perhaps hundreds of scintillation detectors to count the scattered electrons. and the electronics of the scintillation detectors count events and report them versus the scattering angle cosine. such as in the spherical detector illustrated below.Photographic film liner s=R θ R entry port sample s s Scattering of electrons requires more sophistication. 161 . Two colliding beams scatter from the center point of the machine. Out of 6.0 × 106 incident particles (total counts).4 5 3 6 2 1 7 θ 1’ 7’ 2’ 6’ 5’ 3’ 4’ 5 10 Sum to get counter 1 signal counts 104 103 102 101 100 0 50 100 counter 150 Six million particles bombarding total α particles incident on a stationary Au nucleus at the center of the detector produces the following plot of the number of counts dN gotten by the detectors located at angle θ. 3 particles scattered at angles exceeding 169o ! If the gold atom was well described by the Plum-Pudding Model 162 . versus cos θ. there were 1976 particles scattered through angles with θ > π2 = 90o . In the simulated experiment below you will be asked to analyze a similar data set.of Thompsen. and compare it to the theoretical predicted dN (θ) for several models of the atom.0 × 106 ) divided by the cross sectional area of the beam. 17.0 Plotted against this is the theoretical curve obtained from the assumption that there is a nucleus. 105 dN (Ω) 104 103 102 101 100 −1.5 1. namely b db dN (θ) = F dΩ T Ntarget sin θ dθ db dθ T Ntarget = 2π F b dθ in which T is the time of exposure (the time for which the sample was bombarded).0 −0. You can see that the data conforms extremely well with the hypothesis of a concentrated nucleus. This makes F T equal to the total number of projectiles (6.1 The numerical experiment The figure illustrated above is the data analysis of a simulated scattering experiment.0 cos θ 0.5 0. The Rutherford-Geiger-Marsden experiment is a bit too costly to permit its performance on our 163 . this would be completely impossible. Gold layers a few atoms thick can be laid down on a substrate by sputtering or ballistic deposition. and have a radius of about 1 × 10−10 m. with a detector density of one scintillator per degree in the θ direction. We imagine that an apparatus utilizing scintillators such as that on the preceeding page was used. and its atomic mass is 197 mole .0 × 10−10 m. Apparatus. Radium sample Lead shield Collimators The runtime of the experiment is T = 60.93 × 104 m 3 . but we can certainly analyze a generated data set. The data file must then be analyzed to create a histogram or frequency plot of counts versus cosine of the angle at which the event was detected. There is as usual a linear relation between channel n and in this case the angle θ.0 × 10−4 Curies behind a series of collimating aperatures to cut down the the isotropically emitted particles into a tight beam. making dθ = 180 The incident beam. about ten close-packed atoms deep. The density of gold is kg g 1. no individual α is likely to interact with a gold atom with an impact parameter exceeding 1. which tabulates the counts and reports them by channel n.2 Raw data π The actual data is given in the table below. Each scintillator array for angle θ = n 180 is counted by an MCA.0 seconds. my incident beam was made by placing a purified radium sample of 1. but manufacturing techniques have evolved considerably since the time of Geiger and Marsden. 760 eV .0 × 106 events (incident α particles). The target. This is impractically thin. and so Np = 6. π . which writes the number to a π file.0 × 106 . each event triggers the scintillator that detects it to report its number (1 → 180) to a computer. 17. during which time the scintillators record 6. making them deeply nonrelativistic. The target is a very thin gold foil of thickness L = 4. 164 . The α particles are found to have a kinetic energy of 113. In this simulation. Counter n has therefore has angular location θ = n 180 in radians.25 × 10−10 m. Since the gold atoms are close-packed. The machine reports counts in a manner similar to the operation of a scaler-counter.campus. and determine which hypothesis it conforms to. To do so you must first plot the data as dNscatt (θ). db . F · Ntarget = Rp ρNA AL · A M Notice that if we integrate over all scattering angles. 1. versus the cosine of the scattering angle. the number of projectiles scattered into a cone of angle θ to θ + dθ. or a Thompson model atom. You will also need b dθ. You should answer the questions below as well. 2. plot the experimental data against the theoretical curve of dN (θ) versus cos θ. Explain how you can measure the energy of the α particles used in the simulated experiment. 165 . and scattering time. to find each of the following items. target composition and thickness.3 Problem 1. 3. Nscatt = 0 dNscatt (θ) dθ. a nuclear atom. Analysis of simulated lab data You should now work up this data. dθ for the Rutherford model the angular separation of scintillators Next you will need to compute the theoretical number of projectiles scattered into each scintillator using db dNscatt (θ) = 2π F b dθ T Ntarget dθ for the Rutherford nuclear model.17. You should use a log plot on the y-axis because of the tremendous variation in dNscatt with angle. Next you should use the information given regarding beam energy. Nscatt = Np = Rp T = πb2max F · Ntarget T = πb2max Np ρNA AL · T AT M This should be used to verify the supposed value of bmax . since every incident α is scattered by Rπ some angle. Compute the α particle beam rate Rp . Compute the distance of closest approach to a gold nucleus of one of the α particles used in the simulated experiment in a head on collision b = 0. Channel 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 Eα = 113760 eV Counts Channel Counts Channel 474473 42 160 85 3978818 43 153 86 858939 44 137 87 301879 45 137 88 138943 46 139 89 75729 47 105 90 45606 48 126 91 29514 49 96 92 20396 50 116 93 14383 51 81 94 10634 52 91 95 8093 53 72 96 6476 54 77 97 4933 55 73 98 3947 56 76 99 3385 57 70 100 2747 58 65 101 2313 59 50 102 1902 60 52 103 1634 61 47 104 1466 62 53 105 1242 63 51 106 1077 64 34 107 989 65 46 108 893 66 42 109 754 67 29 110 661 68 44 111 584 69 38 112 558 70 36 113 464 71 35 114 432 72 34 115 400 73 42 116 357 74 30 117 314 75 35 118 292 76 29 119 297 77 34 120 281 78 24 121 253 79 26 122 222 80 23 123 184 81 31 124 179 82 21 125 165 83 18 126 166 Counts 16 12 14 15 13 17 10 14 13 16 10 11 10 11 16 8 4 7 12 4 10 7 11 9 10 5 6 8 6 7 5 5 7 8 2 6 7 5 5 3 9 5 Channel 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 152 153 154 155 156 157 158 159 160 161 163 164 165 166 167 169 170 171 Counts 6 6 5 1 3 6 4 5 2 5 5 3 3 8 3 3 3 2 3 3 2 5 3 1 2 1 2 1 2 1 1 1 2 1 1 2 2 1 1 1 1 1 . away. since after the α leaves the atom. it experiences no force. 8000 times its mass. it will be pushed away by the force above. If the incident √ α with impact parameter b < R and velocity v = −v0 i enters the atom at point r(0) = − R2 − b2 i + b j.4 The Thompson model Demonstrate using Gauss’s law that within the Thompson atom.17. the force exerted on an α particle due to the positive charge alone is F= 2Ze2 (xi + yj) 4π0 R3 θ R b R θ 1 The α simply blows the electrons. This is only valid as long as |r(t)| ≤ R. and so travels in a straight line. and so interacts mostly with the positively charged part of the atom. Show that the solutions to the Newtonian equations of motion are √ v0 √ √ √ r(t) = ( √ sinh γ t − R2 − b2 cosh γ t)i + b cosh γ t j γ in which 2Ze2 4π0 mα R3 giving the position vector of the α after entering the atom. Prove that the time at which the α leaves the atom is √ √ 2v0 R2 − b2 tanh( γ t` ) = √ v2 γ(R2 + γ0 ) γ= and demonstrate that the scattering angle is given by √ 2bγ R2 − b2 vy (t` ) = 2 tan θ = vx (t` ) v0 − γ(R2 − 2b2 ) 167 . with the centers of the ellipses being at the center of the atom. so experiments are first designed by simulation before hardware is built or modified. Projectile impacts should be randomly and uniformly distributed. Typically one generates scattering events that obey the dynamics of some theoretical model.5 Monte Carlo scattering simulation One of the most commonly performed numerical experiments today in high energy physics is the Monte Carlo simulation of scattering.5 M eV . as is the case in planetary motion.0 × 10−10 m in radius. The actual Monte Carlo portion of the simulation is the beam generation.6 eV q 1− b2 R2 − (1 − 2b2 ) R2 For the α energy of 113760 eV . rather than the focal points being at the center of the atom. To simulate a beam of spatially uniform flux. Runtime on big experimental hardware is expensive. which is very high. 168 . Show that for the atomic size of R = 1. 17. then for gold (Z = 79) Ze2 = 1137. It was thought in 1912 that the atom was about R = 1. given its estimated size. You can see that this is certainly small enough to reveal details within the atom. and runtime is paid for.0 × 10−10 m. and precise projectile parameters such as impact parameter are simulated using random number generation. consider a phosphorescent screen placed perpendicular to the beam. The final phase is to establish a reasonable value for R. In our simulated experiments we will use much lower energies.0 × 10−10 m. tan θ = 2 Rb Eα 1137. as illustrated below. compute the de’Broglie wavelength.Demonstrate that the electrons embedded in the atom will execute elliptic orbits. You will now run a Monte Carlo for the Thompson model.6 eV 4π0 R Geiger and Marsden used α particles of about 5. If we suppose that R = 1. one√ to act as y. 2. proceed. The resulting impact parameter b = x2 + y 2 is randomly distributed.5 1. 169 .5 0. Generate a uniform random deviate r. 1. √ 2would 2 both between −1 and 1. but not uniformly. Compute which scintillator would detect the scattered α from n = f loor(180 · πθ ) where f loor is the function that rounds a floating point number down to the nearest integer. You could use the inverse method or the accept-reject algorithm instead. Print out this integer. Z a Z πa2 dx dy b db dθ = = ℘[b ≤ a] = 4 4 8 b≤a 0 and so b has an unnormalized PDF of fb (b) = d πb ℘[b0 ≤ b] = 0 db 4 Suppose that Rb is a randomly distributed variable on the interval from 0 to 1 with such a PDF. Write a simple computer program to generate 600.0 0. What is the probability that an α will be scattered by an angle greater than π2 radians? This would require that the denominator be negative. The easiest way be to generate two random deviates x and y. Do this for several α particle energies.0 −0.0 0. and if r = x + y ≤ 1.5 −1. The PDF’s of these numbers are both 21 . 000 scattering events with such a random beam of α particles. one to act as x. 3. Compute the appropriate scattering angle θ from the Thompson scattering formula.0 This uniform beam was simulated by choosing two randomly and uniformly distributed numbers between −1 and 1.0 −0. with a normalized PDF based on that given above.5 0. interpreted as Rb .1.0 −1. otherwise pick two more and try again. One energy can be singled out by passing the beam through a velocity selector. four our beam and target parameters. For each energy for which you have the scattering data. Plot this versus the energy. The radium source emits α particles with a variety of energies. On the next few pages you will find the detector counts for several incident α particle energies. create a frequency plot of the number of counts gotten by each detector. given above. Plot the experimental counts and the theoretical curve on the same figure. determine the number of α particles scattered through more than π2 radians. Beginning with dN (θ) = 2π F Ntarget T b db and θ k cot 2Eα 2 we can determine the energy dependence of the scattering rate on incident energy by counting the number of α particles that should be scattered into angles exceeding π2 radians. In each experiment. Now compute the number of such events predicted by the Rutherford formula. This sub-beam can then be accelerated further with a high voltage.4. we have logged 6.6 Energy dependence in the Geiger-Marsden experiment Geiger and Marsden were very thorough. The data for Eα = 113760 eV was displayed several pages back. crossed electric and magnetic fields. compute χ2 for the four data points. Is the hypothesis verified to be true? 170 . These must be incident on gold nuclei with impact parameters less than b= bmax = k k π cot = 2Eα 4 2Eα and so the total number of scattering events logged by all detectors located at θ > be Z bmax k2 π 2π F Ntarget T b db = π F Ntarget T b2max = π F Ntarget T Nθ> 2 = 4Eα2 0 π 2 should This inverse-energy-squared behaviour is almost universal in high energy physics scattering experiments. Does this look anything like that seen by Geiger and Marsden? 17. They also investigated the dependence of the scattering rate on incident α particle energy.0 × 106 scattering events. Now that you have a data file of detector events. In this problem you will verify the hypothesis that the scattering rate has this behaviour as a function of incident energy. Channel 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 Counts 4453047 1160804 214677 75138 34747 18774 11402 7330 5061 3524 2701 2098 1626 1249 1017 828 692 564 529 438 341 299 274 239 224 207 170 150 122 118 128 99 81 81 70 Eα = 227520 eV Channel Counts Channel 35 67 71 36 61 72 37 55 73 38 60 74 39 54 75 40 49 76 41 50 77 42 52 78 43 42 79 44 35 80 45 39 81 46 35 82 47 25 83 48 28 84 49 21 85 50 18 86 51 16 87 52 18 88 53 19 89 54 16 90 55 15 91 56 18 92 57 11 93 58 15 95 59 11 96 60 14 97 61 16 98 62 5 99 63 8 100 64 7 101 65 8 102 66 9 103 67 10 104 68 11 105 69 9 106 171 Counts 6 6 6 8 5 6 4 4 9 4 3 5 7 2 8 3 1 9 4 5 1 10 1 2 4 4 3 4 2 2 4 3 4 1 4 Channel 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 127 130 131 132 133 134 135 137 138 139 142 143 147 149 151 154 155 159 162 163 Counts 3 2 3 1 2 2 1 2 1 2 2 2 3 1 3 2 1 1 1 1 1 1 1 1 1 2 1 1 2 2 1 1 1 1 1 . Channel 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 Counts 474793 2776956 1202451 556203 302768 183063 118825 81131 57792 42979 32755 25560 20046 16380 13175 11103 9286 7731 6606 5644 4953 4323 3802 3499 2979 2590 2342 2033 1993 1748 1514 1500 1301 1221 1069 988 875 906 789 759 698 664 586 596 Channel 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 Eα = 56880 eV Counts Channel 508 89 520 90 493 91 433 92 406 93 368 94 368 95 334 96 324 97 317 98 267 99 279 100 263 101 238 102 247 103 224 104 204 105 195 106 183 107 195 108 177 109 194 110 163 111 165 112 145 113 147 114 122 115 112 116 125 117 112 118 113 119 119 120 90 121 102 122 100 123 95 124 112 125 81 126 82 127 98 128 66 129 99 130 61 131 68 132 172 Counts 69 58 61 63 51 79 49 56 48 44 45 44 50 35 41 43 39 26 41 40 35 40 33 37 34 37 30 36 29 31 31 23 22 24 13 20 17 15 19 15 18 19 13 15 Channel 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 Counts 16 18 5 10 13 12 15 12 13 11 9 11 6 13 6 11 10 6 11 8 9 6 7 8 7 8 10 5 5 6 5 6 2 1 4 2 2 3 1 4 2 1 2 1 . Channel 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 Counts 476108 1569314 1210099 729550 472849 324276 231905 172040 130800 101969 81012 65076 53825 44150 37007 31211 26514 23014 20010 17451 15256 13624 11972 10545 9500 8568 7665 6969 6444 5712 5277 4840 4427 3981 3712 3411 3199 2947 2759 2522 2343 2326 2025 2066 Channel 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 Eα = 28440 eV Counts Channel 1840 91 1768 92 1597 93 1474 94 1461 95 1320 96 1239 97 1176 98 1166 99 1125 100 1044 101 966 102 885 103 924 104 848 105 822 106 742 107 752 108 681 109 617 110 620 111 565 112 617 113 563 114 538 115 540 116 466 117 493 118 460 119 441 120 400 121 444 122 393 123 350 124 389 125 374 126 339 127 322 128 303 129 272 130 320 131 260 132 260 133 266 134 173 Counts 268 222 236 219 216 214 217 202 175 182 167 175 171 179 185 151 158 153 137 123 125 120 121 118 122 103 120 103 108 111 88 120 81 98 82 83 75 82 68 97 76 60 61 57 Channel 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 Counts 69 52 62 50 55 57 53 56 54 54 53 51 44 37 34 27 29 31 29 28 26 22 22 20 24 22 17 19 13 20 19 13 16 13 16 10 12 7 5 4 4 6 3 1 . the transmitted beam will be bent to angle λ with respect to the normal. and V0 . and the Z electrons are embedded in this ball in such a way that the electrostatic potential outside of the ball is zero. 174 . Consider a simplified atomic model in which the positive charge +Ze is spread out into a uniform ball of radius R.7 Problems 1. requiring that the α beam bend sharply at r = R as illustrated below. Find the scattering angle in the case of 21 mv02 < V0 . Θ s b R φ ψ ψ φ Θ The impact parameter of the bent beam is s. and has kinetic energy exceeding V0 . Perhaps you can see now why Rutherford was so surprised when α particles literally bounced back into his face. a. and inside is V0 > 0.17. The angular momentum will be conserved. Use conservation of energy and angular momentum together with the figure to show that sin φ v1 b = = sin λ v0 s b. Use this to find the scattering angle θ as a function of b. Incoming α particles of impact parameter b and speed v0 will slow down once they get into the ball. to speed v1 . b. Compute the maximum scattering angle as a function of m (α particle mass). If the incoming beam has angle of incidence φ with respect to the normal of the surface of the ball. c. h> #define PI 3. /* Monte Carlo that creates data for Thompson atom scattering */ #include <stdlib. /* constants and the energy */ E=11000.b. 0. Physically counting in each case we find 116. for(n=0. determine the rate at which particles are scattered through angles greater than 0.n++){ 175 . so that the simulations are quit good. dΩ This 3.1415926 int n.0 M eV alpha particles incident on a gold foil of thickness 1 × 10−6 m. int crap.182 × 10−26 Rc2 60 s (0.2*. 471.den.023 × 1023 )(4.02.93 × 104 )(6.1 radian.0.h> #include <stdlib.8 Notes The data was generated using energies Eα = k Rc with Rc = 0.m2. find the differential cross section is a slightly more challenging problem.25 × 10−10 ) Rc2 )(60 s) = 1. E0=1137.0). Maximum impact parameter was in practicality around 1 × 10−10 m.31.6. 1890. 473.R_cl. For the case 21 mv02 > V0 in the first problem.04 0. dσ .num. establish MAX=2^{31} */ srand(17).E0.theta. For a stream of 6.h> #include <math. 0. main(){ float m1. max=pow(2. 17.test.det. 1885 and 7540.n<600000. /* seed the random number generator. and 7563 α particles scattered by more than π2 radians for Rc = 0. E. This makes πF Ntarget T π( k2 = 4Eα2 6 × 106 (1.2 × 10−10 meters for the 113760 electron volt α.0.197) 4 which will give 118.max.08 × 10−10 m which gives the four energies used to generated the energy dependence data.01. 0*b*b).023 × 10 )(4.62) sin θ π sin4 θ2 180 #include<stdio.0*b*sqrt(1. 176 .02 × 10−10 cos 2θ sin θ = 2 sin4 θ2 sin3 θ2 db dθ F T Ntarget dθ (1.93 × 104 )(6.n<180.02 × 10−10 )2 sin θ ) π 6 (1.0*theta/PI).den). } } Below is the program to generate data for the theoretical Rutherford scattering. /* print out detector logging event */ det=floor(180. b=sqrt(m1*m1+m2*m2). den=(E/E0)-(1.h> #include<math.25 × 10−10 ) π · 6 × 106 · (0.0.197) 180 2πb = 2π · ( k k 4 0. det).0.sig. m2=2.197) 180 sin 2 = 2π (37.h> #define PI 3.theta. sig=sig*sig.93 × 10 )(6.1415926 int n. if(b>1.0*(float)rand()/max-1.25 × 10 · 6 × 10 · 4 θ 4·4 (0.0)./* here is the Monte Carlo step */ generate:{ m1=2.0) goto generate. for(n=1.n++){ theta=(float)n*dtheta*PI/180.0-b*b).0*(float)rand()/max-1.0.0-2. theta=atan2(num.02×10 −10 = 2π 1 θ 1 )2 cot 2 2 sin2 θ 2 4 23 −10 (0.0. sig=sin(theta/2.023 × 1023 )(4. /* now create the event */ num=2. float dtheta. main(){ dtheta=1. using E= k 0. printf("%d\n". we should come up close to six million.0) goto generate. printf("%d\n". /* creates data for rutherford scattering */ #include <stdlib.0*(float)rand()/max-1. theta).0*PI*(37. det).n++){ generate:{ m1=2.0*(float)rand()/max-1.max.0*theta/PI).h> #define PI 3. cos(theta). for(n=0.0.62)*sin(theta)*(PI)/(180. printf("%f\t%f\n". int crap.0*sig).n<6000000.0). } } Summing over all output values. */ /* modify to print out detector number */ det=floor(180.31.sig=sig*sig.1415926 int n.m2. R_cl=0. sig=2.test.h> #include <math. } } 177 .02. b=sqrt(m1*m1+m2*m2). srand(17).0*b).det.0. The Monte Carlo program to create the “experimental data” is below.R_cl.2.0. main(){ float m1. theta=2. m2=2.theta.0*atan2(R_cl. /*printf("%f\n".h> #include <stdlib.} if(b>1. max=pow(2.b.sig). sort -n < list > sorted_list which looks like this 178 . and it is difficult for you to use the support programs included in the manual. Suppose that we count nuclear decays for 10 second periods.1 Appendix Appendix I. and there are lots of Awk scripts strewn throughout the manual. but what if we had hundreds? We can sort the data using the Unix program sort. try to use Awk to process your data. with the −n switch telling sort to sort by arithmetic value. Awk is programmable and uses the syntax of C and shell programming. It is easy to learn. and is part of any Linux installation. so the command cat list produces the output above. one number per line as shown above.18 18. We need to find the largest and smallest values. Awk is a free download for Win32 machines from the physics department website. easy for 20 data points. obtaining the following 20 measurements 1016 1207 1186 1244 1110 1099 1099 1185 1220 1286 1117 1280 1190 1083 1177 1189 1200 1201 1188 1291 which we enter into a file called list . As an example we will process data for the Nuclear Counting Statistics experiment. If you have little or no programming experience. Reformatting Lab Data with Sed and Awk Sed and awk are two Unix editting tools that are particularly useful in processing data taken in lab. treating each line as a record with several data fields. beginning with a null character. $2. one record per line. RS = ""} { print "smallest is " . the field separator FS is a newline character. Here is the awk script min max. " " . The normal operation mode of awk is that awk acts on each line of a file in turn. $NF} We can obtain the largest and smallest numbers in our data set by running sort on the file list and piping the output into this awk script sort < list | awk -f min_max. The record separator RS will be a blank line.1016 1083 1099 1099 1110 1117 1177 1185 1186 1188 1189 1190 1200 1201 1207 1220 1244 1280 1286 1291 This file will have a first line that is blank. a feature of some versions of sort. each separated by a field separator symbol that is user defined. this means that the second line of the file sorted list and the last lineare the smallest and largest entries in our data file. and so forth. $2. The last data field is called $NF.awk that will read the file and report the second and last records to the console. Smallest is 1016 largest is 1291 179 .awk BEGIN { FS = "\n". Check for this behaviour when running sort for the first time .We can pipe the output of this program into an awk script min max.awk which produces the output line .awk #min_max. We need to tell awk that it will read a file whose records are single lines in column format. The fields are accessed by refereing to them by the names $1. " largest is". $1 ++i} } This will run through the loop for each line of the file list in turn. The range of counts is Nmax − Nmin = 1291 − 1016 = 275 We can sort our data into 11 bins of width 25 counts by running the data in list through the following awk script called bin sort.The next stage in processing the data for this experiment is to sort the data into bins of a given width in count-space. assign each data point to a bin. The output of awk -f bin_sort. and print the bin first.awk #bin_sort. " ".awk list is 0 7 6 9 3 3 3 6 8 10 4 10 6 2 6 6 7 7 6 10 1016 1207 1186 1244 1110 1099 1099 1185 1220 1286 1117 1280 1190 1083 1177 1189 1200 1201 1188 1291 We can sort this data by running 180 .awk { i=0 while ( i<11 ) { if ( $1 >= 1016+25*i && $1 <= 1016+25+25*i ) print i. then the data point separated by a blank space. awk list | sort -n | uniq -c which produces output 181 . Note that sort will perform numeric sorting based on the first field of each data line.awk { i=0 while ( i<11 ) { if ( $1 >= 1016+25*i && $1 <= 1016+25+25*i ) print i ++i} } We run the command awk -f bin_sort. bin population followed by bin number. 0 2 3 3 3 4 6 6 6 6 6 6 7 7 7 8 9 10 10 10 1016 1083 1099 1099 1110 1117 1177 1185 1186 1188 1189 1190 1200 1201 1207 1220 1244 1280 1286 1291 We can do even better by telling bin sort.awk list | sort -n which will produce the sorted output below. #bin_sort.awk to print only the bin number of each data point and piping the output through sort into the uniq command that will list the number of occurances of each bin label in the list.awk -f bin_sort. and output two columns. "\\\\". \begin{tabular}{|c|c|}\hline Bin label & Bin population \\ \hline 0 & 1 \\ \hline 2 & 1 \\ \hline 3 & 3 \\ \hline 4 & 1 \\ \hline 6 & 6 \\ \hline 7 & 3 \\ \hline 8 & 1 \\ \hline 9 & 1 \\ \hline 10 & 3 \\ \hline \end{tabular} 182 .1 1 3 1 6 3 1 1 3 0 2 3 4 6 7 8 9 10 We can save this data to a file called bin pop with awk -f bin_sort. "\\hline"}’ bin_pop which produces 0 & 1 \\ \hline 2 & 1 \\ \hline 3 & 3 \\ \hline 4 & 1 \\ \hline 6 & 6 \\ \hline 7 & 3 \\ \hline 8 & 1 \\ \hline 9 & 1 \\ \hline 10 & 3 \\ \hline We now add a few lines to this and we have our processed data in the form of a nice LaTeX table. $1.awk list | sort -n | uniq -c >bin_pop and print out the contents of the file in a nice format that can be imported into a LaTeX lab report as a table awk ’{ print $2. "&". bin bin_pop using the script #table. "&". 183 . "\\hline"} which produces the output 0 & 1 & 1028 \\ \hline 2 & 1 & 1078 \\ \hline 3 & 3 & 1103 \\ \hline 4 & 1 & 1128 \\ \hline 6 & 6 & 1178 \\ \hline 7 & 3 & 1203 \\ \hline 8 & 1 & 1228 \\ \hline 9 & 1 & 1253 \\ \hline 10 & 3 & 1278 \\ \hline After adding afew LaTeX lines this becomes the table Bin Bin pop.and this prints as seen below Bin label Bin population 0 1 2 1 3 3 4 1 6 6 7 3 8 1 9 1 10 3 We can include another column for the center of the bin with awk -f table. "\\\\".awk { print $2. "&". 1016+12+25*$2. $1. Bin center 0 1 1028 2 1 1078 3 3 1103 4 1 1128 6 6 1178 7 3 1203 8 1 1228 9 1 1253 10 3 1278 Awk and sed can not only be used to easily process huge data files for analysis. but also to prepare data processed into convenient reports. called nonlinear least squares regression. This can be done by finding the best fit to such a function using a minimization process. Suppose that the theoretical count number versus θ is a function N (θ) = a0 + a1 cos θ + a2 cos2 θ or that we wish to fit our data to such a theoretical curve. in which the number of counts gotten by a scintillation counter placed at position θ1 is x¯1 (θ1 ) ± σ1 . which will compute the parameters describing the theoretical form most closely agreeing with our data out of all possible theoretical curves in the same family. · · · N } taken in the lab.18. for example consider scattering data. Nonlinear Least Squares error matrix method Suppose that we have a set of data points with average values and standard deviations {¯ xi ± σi |I = 1. Maximizing with respect to the parameters {ai } leads to the equations X i (¯ xi − ξi ) ∂ξi =0 σi2 ∂am 184 . Suppose that the likelihood of x¯i being the measured value is given by a Gaussian distribution L= 1 q 2πσi2 e − (¯ xi −ξi )2 2σ 2 i then the probability that the entire set of experimental data actually being found to have the values gotten in the experiment will be L= 1 N (2π) 2 σ1 · · · σN e − PN i=1 (¯ xi −ξi )2 2σi We can maximize this probability with respect to the parameters ai . These are graphed with error bars versus a parameter upon which they depend. Our experimental data is of the form x¯i (θi ) ± σi The theoretical values corresponding to these experimental values will be ξi = a0 + a1 cos θi + a2 cos2 θi and in general have the form ξi = n X Cim am m=1 if there are n constants ai involved in the theoretical curve expression. and must find the coefficients ai .2 Appendix II. 2. and we obtain a set of average values for these parameters ¯ a. What is the standard deviation in the values of these fit parameters? (am − a ¯m ) = X k ¯k ) = (M −1 )mk (Xk − X X (M −1 )mk kj Cjk (xj − x¯j ) σj2 however if we assume that the measurements resulting in each data point are uncorrelated. Example Consider the following data table 185 .or X i X Cim Cil Cim x¯i = al 2 σi σi2 i. then < (xj − x¯j )(xi − x¯i ) >= σi2 δij and so < (am − a¯m )(al − a ¯l ) >= = X X (M −1 )mk ijkp Cjk Cip (M −1 )lp 2 σi2 δij 2 σj σi (M −1 )mk (M −1 )lp Mpk = (M −1 )ml kp and so the standard deviation in am is σam = q 1 (M −1 )mm So by finding the matrix M we essentially find the curve of best fit and its tolerances.l We can define the vector of experimental results Xm = X i and the error matrix Mml = X i Cim x¯i σi2 Cim Cil σi2 and then we find X = Ma which has solution a = M−1 X Suppose that hundreds of identical experiments have given us the opportunity to compute the set of parameters a many times. 1 + + + = 3200 0.9 0.01 0.01 0.01 0.1 We propose a theoretical form f (x) = a0 + a1 x2 The theoretical values of f(x) should be then ξ1 = a 0 ξ2 = a 0 + a 1 ξ3 = a0 + 4a1 and ξ4 = a0 + 9a1 We find the matrix Cim is     Cim =  the data vector elements are X1 = 1 1 1 1 0 1 4 9      2.01 and the error matrix is Mlm = This has inverse (M  0.1 0.1 1.0007143 0.9 8.9 8.0002041 ! 100 100 100 100 0 100 400 900 = −1 0 1 4 9 )lm = 400 1400 1400 9800 186 !      1 1 1 1     ! .1 1 2.1 +4 +9 = 21040 0.01 X2 = 2.9 19.01 0.1 3 19.005 −0.x 0 f(x) 1.01 0.0007143 −0.1 2 8.9 19.1 σ 0.9 0. 0007143 0. such as those needed to limit the abcissa and ordinates xmin ≤ x ≤ xmax.We arrive at the following parameters a1 a2 ! = 0.01428 are the errors in the determination of these best fit parameters.3 Appendix III. Furthermore we know √ √ σa1 = 0. For example graph -T X data will output the resulting figure into xplot. Postscript. one just passes this choice to graph at invocation and redirects the resulting data into a file.ps or plotting several data files on the same figure graph -T ps data1 data2 data3 > data.info file for more command line options.071.005 = 0. xfig. To output to pnm. ode and several others. for example graph -T ps data > data.ps or perhaps using a logarithmic x-axis within domain 100 ≤ x ≤ 104 graph -T ps -l x -x 0 4 data > data. The Plotting of Lab Data with GNU plotutils This suite of graphics utilities can be used to build publication quality graphs from raw data.9711 + 2. σa2 = 0. It consists of the command-line programs graph. The use of graph is the subject of this section.000204 = 0.0085 ! and so the function f (x) = 0.ps The labels can have Greek letters or math symbols. or GIF. spline.ps Read through the voluminous /usr/local/info/plotutils.005 −0. chosen with the commandline option -T. which are given later in a table in this document 187 .9711 2. Graph can output is work into a variety of formats and devices. plot. 18. m’’ -Y ‘‘Volts’’ -L ‘‘Voltage per meter’’ data > data.0002041 ! 3200 21040 ! = 0.0085x2 best fits our lab data. ymin ≤ y ≤ ymax graph -T ps -x xmin xmax -y ymin ymax data > data. an X windows primitive graphing canvas.ps One can add labels to the x and y axes as well as a banner label graph -T ps -X ‘‘ x.0007143 −0. 15" 15 --reposition 0.3 -L "a=0.11 500 400 400 300 400 300 300 200 200 200 0 −15 100 100 100 −10 −5 0 5 10 15 0 −15 −10 −5 a=0.0 0.t) for t=0.6 0.09" 09 --reposition 0.3 -L "a=0. or to make multigraphs consisting of many repositioned plots.17" 17 --reposition 0.3 -L "a=0.0 0.0 0.07" 07 --reposition 0.0 0.6 0.23 400 400 300 300 200 200 100 100 400 300 200 100 0 −6 −4 −2 0 2 4 6 0 −6 −4 −2 0 2 4 6 0 −6 −4 −2 0 The fact that these programs are command-line means that they can be used in shell-scripts or batch files to perform bulk processing of graphics files.3 0.graph -T ps -L ‘‘ \*Q\sp*\ep(x.3 -L "a=0.3 -L "a=0.6 0.6 0.15 500 500 400 400 300 300 200 200 100 100 5 10 a=0.0 0.19" 19 --reposition 0.3 0.ps will print “Ψ∗ (x.21 500 0 2 4 6 2 4 6 a=0.23" 23 >page1.3 0. For plotting discrete data points the switch −S allows you to use a large variety of symbols for individual data points 188 .17 400 300 200 100 0 −8 −6 −4 −2 0 2 4 6 8 0 −8 −6 −4 −2 a=0. For example the command graph -T ps --reposition 0.13" 13 --reposition 0.3 -L "a=0.13 0 5 10 15 0 −10 −5 0 a=0. Graph has the capacity to produce graphs within graphs.t)\*Q(x.07 a=0.1s” along the top of the figure.3 0. t)Ψ(x.3 -L "a=0. t)f ort = 0.11" 11 --reposition 0.3 -L "a=0.3 0.3 0.21" 21 --reposition 0.19 0 2 4 6 8 0 −6 −4 −2 a=0.6 0.0 0.ps will create the following multiplot a=0.6 0.3 -L "a=0.1 s’’ data > data.09 700 600 600 500 500 a=0. Symbol number 1 2 3 4 5 Symbol · + * × with 31 different symbols in all.01 data > data. For example graph -T ps -S 4 0.0 0 1 2 3 4 5 6 or we can try graph -T ps -C -m 3 data1 --reposition 0. 270 degrees with –rotation graph -T ps --rotation 90 > data.5 2 1 0 0 1 2 3 4 5 6 0.ps Lines connecting data points in plots can be manipulated with -m.5 −1. line mode 0 is no connecting line.01 times the size of the box within which the entire plot is drawn.ps results in 1.5 0. and even color can be manipulated.0 5 4 3 0.5 0.5 0. We can place a plot within a plot by repositioning the second plot within the first and shrinking it. width.0 −0. 180.3 data2>data.1 data2>data. graph -T ps -C -m 2 data1 --reposition 0. Line style. Plots can be rotated by 90.ps which produces 189 .5 0. but adding -C changes this to color linemode rather than breaking the line in different ways.ps uses circles whose size is 0. For example we use -m to change linemode from solid to dotted or dashed. 0. A partial table of useable symbols is given below symbol α ∆ µ Θ Υ ζ θ ψ ± ∪ H ∅ 6 = escape seq. 0. \∗b \∗e \∗n \∗R \∗W \∗d \∗p \if \pt \gr \hb \c+ \ >= symbol Γ Φ η Σ Ξ γ φ ← ∂ P † ⊗ ≤ 190 escape seq.0) being the upper right corner. \∗G χ \∗x \∗F Λ \∗L \∗y Π \∗P \∗S τ \∗t \∗C Ψ \∗Q \∗g σ \∗s \∗f λ \∗l \<− → \− > \pd ∩ \ca R \SU \is \dg ℘ \wp \c∗ ℵ \Ah √ \ <= \sr .5). and (1.5. These include Greek letters and mathematical symbols.5 5 4 3 2 1 0 0 1 2 3 4 5 6 0. symbol \∗a β \∗D  \∗m ν \∗H ρ \∗U Ω \∗z δ \∗h π \∗q ∞ \+− ∝ \cu ∇ \li h ¯ \es ⊕ \! = ≥ escape seq.5 −1.3 will be place with its lower left corner at (0. One of the most useful features of libplot is the generous collection of font symbols that it recognizes for drawing strings on plots.0 −0. In the first example a miniature virtual display of size 0. These few examples are enough to get started with in using graph. For example graph -T X -L "P\sb2\eb(cos\*h)" will result in the string P2 (cos θ) being drawn on the plot title or label.0.0 0 1 2 3 4 5 6 The –reposition is relative to the point (0. 0.0 0. and you really should read through the info file to learn more since this barely scratches the surface.0) being the lower left corner of the physical display.1. 1. symbol escape seq. \\*h)" writes Ψ∗n. and the data analysis programs will output LaTeX tables and formatted equations. The basic math equations are typeset in math mode in which offset. and it excels at these tasks. The support software for this manual contains templates for all of the labs.l. Subscripts and superscripts can be created using the delimiters illustrated by the following example. 18.\*f. $$ \int_{-a}^{\infty} f(x) dx= F(a)$$ Z ∞ −a f (x)dx = F (a) $$ \oint_C {f(z) \over z-z_0 } dz = 2 \pi i f(z_0) $$ I C f (z) dz = 2πif (z0 ) z − z0 $$ \vec{A} \times \vec{B} = \vec{C} $$ ~×B ~ =C ~ A $$\vec{\nabla} \cdot \vec{E} = {\rho \over \epsilon_0} ~ ·E ~ = ρ ∇ 0 191 $$ . It is installed on several of our lab computers. φ. LaTeX is designed to typeset equations and tables.l.Some of these symbols are only available if pl fontname has been used to set the font to Hershey or HersheySans. graph -T ps -L" L\sp\*a\ep\sb n\eb(r)" > out.ps writes Lαn (r) to the label.m\eb(r. and so below you will see the LaTeX commands and corresponding typeset equations for a wide variety of mathematics. the math mode delimiters. This PostScript output is being redirected to a file. In addition it is free software available in binary format for all existing operating systems. The language is best learned by example. Nearly every scientific textbook and journal is typeset in TeX or LaTeX. graph -T X -X"\*Q\sp*\ep\\sbn. LaTeX has the further advantage of being platform independent. θ) to the x-axis label.m (r.4 Appendix IV. centered equations are surrounded with double dollar signs. Preparing a lab report with LaTeX LaTeX is very easy to learn and produces documents of superior quality. $$ \gamma \rightarrow \beta \bar{\beta} $$ γ → β β¯ $$ \oint \vec{p} \cdor d\vec{q}= 2 \pi n \hbar $$ I p~ · d~q = 2πn¯ h $$ \sum_{n=0}^{\infty} {1 \over n^s} = \zeta{s} $$ ∞ X 1 = ζs s n=0 n $$ \prod_{n=1}^{\infty} (1-{x^2 \over n^2 \pi^2})={\sin x \over x} $$ ∞ Y n=1 (1 − x2 sin x )= 2 2 n π x $$ \lim_{n \rightarrow \infty} (1-{x \over n})^n = e^{-x}$$ lim (1 − n→∞ x n ) = e−x n $$ P(x)={1 \over \sqrt{2 \pi \sigma^2}}e^{-{(x-\bar{x})^2 \over 2\sigma^2}}$$ P (x) = √ (x−¯ x)2 1 e− 2σ2 2πσ 2 $$ \ln |x| = \int_{1}^{x} {dy \over y}$$ ln |x| = $$ \alpha \beta \gamma \delta \pi Z x 1 dy y \lambda \epsilon \sigma \omega $$ αβγδπλσω $$ 1={2 \over \pi}[\sin x +{1 \over 3}\sin 3x +{1 \over 5}\sin 5x +\cdots] $$ 192 . Kenosha } University of Wisconsin-Parkside \\ WI \\ University of Wisconsin-Parkside \\ WI \\ \begin{abstract} Here is the very short description of your experiment. Kenosha \and author 2 \\ Department of Physics. Type this into a file to use as a template for all of your lab reports. \end{abstract} \maketitle \noindent {\bf Section 1. It can be included in the document with the code line \\ \rotatebox{270}{\scalebox{0. 193 . 1= Below is the basic template for the LaTeX documentstyle article.5]{\includegraphics{apparatus. draw a diagram of the apparatus using xfig and convert it to a postscript figure. 900 Wood Road.1 1 2 [sin x + sin 3x + sin 5x + · · ·] π 3 5 This shows a pretty full range of LaTeX math typesetting commands. 900 Wood Road.}\\ \\ Describe the experimental procedure.ps}}}\\ \noindent {\bf Procedure.}\\ \\ Describe the equipment used.5}[0. \documentclass[12pt]{article} \pagenumbering{arabic} \usepackage[dvips]{graphics} \begin{document} \title{ Your Title Here } \author{ author 1 \\ Department of Physics. There is a handout on using awk and sed to process your data files and put them into LaTeX table format. Any graphs generated can be included in this section with \rotatebox{270}{\scalebox{0. This is the place where your error analysis and propagation is performed and your results compared to the accepted (correct) values. \noindent {\bf Data.Be to the point. Here is a typical 4 column table including some multicolumn rows.5}[0. \begin{tabular}{|c|c|c|c|}\hline \multicolumn{2}{|c|}{\bf Trial } & \multicolumn{2}{|c|}{\bf Trial 2}\\ \hline Bin & Number & Bin & Number \\ \hline 1 & 3 & 5 & 9 \\ \hline 2 & 4 & 6 & 3 \\ \hline 3 & 7 & 7 & 1 \\ \hline 4 & 1 & 8 & 2 \\ \hline \end{tabular} Notice that an ampersand is used to separate the table fields and a LaTeX newline \\ ends each line in the table. \noindent {\bf Conclusions..}\\ 194 ./ps/graph. brief and go step by step.}\\ \\ This is the section in which you perform sample computations and analyse your data. \noindent {\bf Analysis.5]{\includegraphics{.ps}}}\\ It may not be necessary to include the rotate box command here.}\\ \\ Your data must be in tabular form. For External triggering.tex If there are no errors.\\ In this section draw your conclusions. An excellent book is A Guide to LaT eX2 by Helmut Kopka and Patrick W. Horizontal display is controlled by the triggering sections on the face of the scope. For Internal triggering. Line setting triggers from the 60 Hz power line that powers the scope. Daly.dvi. type it up using this model and call it report. the wave being observed triggers itself. sweep is initiated whenever a preset voltage level is reached. It wiil prompt you. The Oscilloscope This is a device used to analyse electronic waveforms and signals. If there are errors the LaTeX engine will stop on the offending line and point out the error. easily remedied.ps which can be printed.tex. second edition 1995. It is well worth your while to become adept with LaTeX if you plan on pursuing Physics or any technical field. Trigger signal is the same as the input vertical signal. Addison-Wesley. \end{document} To format your lab report. As this voltage 195 .dvi The result of this is report. applied to the Trigger in terminals. Convert this to postscript with dvips -o report. in this case by an externally supplied source. Open your report in the editor and fix the error.ps report. respond by typing X which exits LaTeX. you will be left with a new file called report. Run LaTeX on it from the command line in the directory the file lives in latex report. To measure a signal one does the following. Apply the signal to the Input terminals of the vertical deflection amplifier. Incidentally. 2. 1. Usual LaTeX errors are unpaired dollar signs and other simple things.5 Trial 2 Bin Number 5 9 6 3 7 1 8 2 Appendix V. Auto mode means that the scopes internal sawtooth generator is used to initiate the sweep. what does the table example in the lab template look like? Bin 1 2 3 4 Trial Number 3 4 7 1 18. 1 sec Vert. Horizontal sweep rate can be set with the Sweep time control cm 4. The y(x) relationship between two applied voltages y(t). 196 . Pos. Pos.2 0.c” will generate χ2 in tabular format extremely quickly using the methods of Gaussian quadrature. then the sweep starts over.increases.1 Scale illum. It is limited to no more than about 25 degrees of freedom. and an AC or DC portion of the input signal.1 Ext V/cm 2 5 10 20 10 5 µ sec 1 0.2 1 0.6 5 10 2 10 20 50 1 5 mV 1 0. Appendix VI. This bypasses the sawtooth generator and horizontal deflection is controlled by x(t) If the two input signals are harmonic with frequencies in a rational ratio. msec 2 AC off Vertical V/cm Input 1 0.5 0.5 0.5 0. x(t) can be determined by setting Horizontal sweep to External and applying x(t) to the Horizontal input terminal and y(t) to the Vertical input terminal. 3. 50 20 5 10 18. a Lissajous figure will be displayed on the screen. the voltage at which the sweep is initiated is set by the Stability level control. the horizontal trace moves from left to right on the screen until a preset value is reached. which can select the voltage with a positive or negative slope. Support software The program below “chi2 table gen.1 input Horiz. Trigger Mode Line Int Auto Ext Trigger in Dc - Stability + AC Focus Intensity + Horizontal sweep time/cm 0. In all but Auto modes. high).n++){ chi=low+(float)n*dchi.n. printf("\\begin{tabular}{|c|c|}\\hline\n"). exit(0). 197 . high is upper bound (floats).0){ printf("Invalid lower bound\n").number. printf("%f & %f\\\\\n". printf(" $\\chi^2_d$ & $\\wp(\\chi^2\\ge\\chi^2_d)$\\\\\n"). P=chi2function(chi. printf("low is lower bound.\n"). number=atoi(argv[3]). } low=atof(argv[1]). char *argv[]){ if(argc < 5 || argc > 5){ printf("chi2_table_gen low high number degree_of_freedom \n").P). exit(0).P.chi. DF=atoi(argv[4]). low. float chi2function( float aa. printf("Generating $\\chi^2$ probabilities $\\{1\\over 2^{{%d\\over 2}} \\Gamma({%d\\ov printf("for $%f\\le x \\le %f$\n\n".#include<stdio. exit(0).} if(DF<1 || DF > 20){ printf("Invalid number of degrees of freedom\n").\n"). if(low<=0. printf("\\hline\n").h> float gammaf(float x). printf("between low and high. printf("number is how many table entries to generate\n").h> #include<plot.} if(high < low){ printf("Invalid upper bound\n").n<=number. for(n=0. int DOF). int DF.} dchi=(high-low)/(float)number.chi.h> #include<math. exit(0).DF). printf("degree_of_freedom is an integer <= 20.h> #include<stdlib.dchi. float high. low. high=atoi(argv[2]). main(int argc.\n"). y[5]=5.13028248.56591623. y[13]=38. } The program requires the following (called “chi2.02608557.63027578e-1. y[2]=1. Uses Laguerre polynomial roots */ /* accuracy is limited to 4-5 decimal places */ /* use GMP lib to improve the accuracy */ #include<stdio. /* the weight factors */ w[0]=2.42533663. y[10]=20.18234886e-1.65440771. y[14]=48.denominator. 198 . w[1]=3. float chi2function( float aa.h> #define PI 3. float y[15]. /* computation of chi^2 statistic by Gaussian */ /* quadrature.26994953.exponent. N= degrees of freedom */ /* good up to N=14 but is best well below that point */ int m. /* the abscissa points */ y[0]=9.926917403e-1.62389423.66762272. y[9]=16.40751917.12022856.42210178e-1. y[7]=10. y[6]=7.53068331. w[2]=2. y[1]=4. y[4]=3.1415926 float gammaf(float x).} printf("\\end{tabular}\n").215595412. float sum. int DOF){ /* a=experimental chi^2. y[12]=31.c” ) subroutine to compute the perform the quadrature. float w[15]. y[3]=2. y[11]=25. y[8]=13.77647890.p.3307812017e-2.h> #include<math. m<15.193527818.0-1. w[4]=4.00. b[3]=-0. denominator=pow(2. } return(exp(-aa/2.21243614e-3. b[1]=-0.45651526e-13. exponent ). w[5]=8.56387780e-3. w[10]=4. 199 . prefactor. sum=0. w[6]=1.w[3]=1. w[12]=1. } Also required is the following rather inefficient program for the gamma function “gammaf. exponent=(float)DOF/2. int m.0. float b[9].22631691e-7. w[7]=1.756704078.60059491e-20.918206857. b[8]=0. w[13]=1.45992676e-6.11674392e-4.arg[9].h> #define PI 3.92189727e-11. w[8]=6.m++){ sum=sum+w[m]*pow(2.0. w[9]=2.897056937.1415926 float gammaf(float x) { /* The Gamma function */ float sum.0*y[m]+aa.48302705e-16. b[0]=1.c” #include<math.577191652. b[2]=0.02068649e-2.22743038e-9.26425818e-1. w[11]=3. for(m=0. b[6]=0. b[7]=-0. exponent)*gammaf((float)DOF/2.0).0)*sum/denominator).988205891.0.482199394. b[5]=-0.035868343. w[14]=1. b[4]=0. } if(x>0.0){ do{ x=x-1.0.0.0.0.0 && x<1.c -o chi2_table_gen 200 -lm .} return(prefactor*sum). x=x+1.0) /* this is the only recursive call (1) */ return(PI/(sin(PI*x)*gammaf(1-x))). if(x>2.} while(x>2.} for( m=1.sum=1. sum=sum+b[m]*arg[m]. prefactor=prefactor*x. if(x<0.m<=8.m++){ arg[m]=(x-1)*arg[m-1]. } everything can be compiled with the command gcc chi2.0.c gammaf. arg[0]=1.c chi2_table_gen.0).0){ prefactor=1.0/x. prefactor=1. 189 linetype. 126 conventional cell. 83 Geiger counter. 111 dipole selection rule. 111 equipartition theorem. 36 neutrino. 122 ionization potential. 187 four-momentum. 28 interpolaion. 22 adiabatic process. 20 Newton’s method. 24 fcc structure. 6 chi squared gen. 20 Binomial theorem. 43 longitudinal waves. 36 Hershey fonts. 13. 126 Franck-Hertz experiment. 55. 24 Bargmann-Fock space. 6 isotopes. 127 Compton scattering. 53 Bravais lattice. 189 lissojou figure. 190 oscilloscope. 61 bulk modulus. 89 Bragg condition. 62 Laue condition. 85 201 .Index accelerating voltage. 80 Normal distribution. 41 half-life. 47 density of states. 57. 40 alpha particle. 25 graph. 63 Bragg scattering. 112 escape sequences. 145 Neon. 143 fig. 148 coincidence circuit. 36 null measurement. 43. 62 angular correlation. 49 hypergeometric function. 148 Compton edge. 40 markers. 63 linemode. 5 K capture. 143 bcc structure. 191 high pressure gas tank. 188 Miller indices. 16 EM wave. 41 binsort. 13 nuclear spin. 22 beta decay.c. 36 Curie. 53 accept-reject. 63 beam-splitter. 47. 42 de Broglie relation. 22 lattice spacing. 16 Cobalt-60. 145 Fermis’ Golden Rule. 20 Geiger tube. 63 molecular weight. 126 de’Broglie wavelength. 63 Coulomb energy. 175 activity. 20 Aluminum crystal. 187 gnuplot.c. 83 ionizing radiation. 63 Fermi Golden Rule. 21 Helium-Neon laser. 56 degree of freedom. 20 GIF. 36 characteristic curve. 187 background radiation correction. 16 Bohr levels. 6. 5. 37 Bequeral. 145 angular momentum. 36 Awk script. 41 speed of sound. 89 scaler-counter. 69 Planck apparatus. 13. 5 radio-isotope generator. 70 202 . 61 specific heat. 20 scattering planes. 53 selection rules. 187 primitive cell. 61 unbiased estimator. 8 polarization. 175 Sodium. 23 Rayleigh-Jeans law. 122 translation vectors. 188 rotational degrees of freedom. 112 reposition. 69 Planck radiation law.photoelectric effect. 71 symbols. 114 work function. 150. 187 Poisson distribution. 36 space lattice. 62 radiation. 145 simulation. 17 Poissonian distribution. 114 pnm. 45 Rydberg constant. 41 Stefan-Boltzmann law. 5 scaler-timer. 114 stopping voltage. 65. 145 Postscript. 17 wave quation. 190 T versus R program. 111 Wein’s law.