Unit 12A Quantum PhysicsDue: 7:30am on Friday, March 27, 2015 To understand how points are awarded, read the Grading Policy for this assignment. Prelecture Video: Photons (Chapter 28, Video 1) Click the Play button below to start the video. You must answer the questions embedded in the video in order to proceed through it. When you have watched the entire video, answer the followup question below in Part A. Only this followup question will be graded. Part A In the video, we see that sunscreen protects the skin by ANSWER: Reflecting ultraviolet light. Absorbing ultraviolet light. Shifting the photon energy of ultraviolet light to a safer range. Healing the damage caused by ultraviolet light. Correct Prelecture Video: Energy levels and quantum jumps (Chapter 28, Video 3) Click the Play button below to start the video. You must answer the questions embedded in the video in order to proceed through it. When you have watched the entire video, answer the followup question below in Part A. Only this followup question will be graded. Part A The spectrum of light emitted by helium atoms is different than that emitted by hydrogen atoms because ANSWER: . The force binding an electron to a helium nucleus is larger. You must answer the questions embedded in the video in order to proceed through it. answer the followup question below in Part A. When you have watched the entire video. The different atoms have different quantized energy levels. Correct Prelecture Video: The wave nature of matter (Chapter 28. . so move more slowly. and hydrogen is not.Helium is a noble gas. Only this followup question will be graded. Helium atoms are more massive. Video 2) Click the Play button below to start the video. Particles have a wave nature as well.Part A Light has both a wave and a particle nature. and therefore ANSWER: . 63×10−34J ⋅ s for Planck's constant and c = 3. Express your answer in electron volts.63×10−34J ⋅ s is Planck's constant.00×108m/s is the speed of light. f is the frequency. ANSWER: .00×108m/s for the speed of light and express your answer in electron volts. where 6. the maximum kinetic energy of the emitted photoelectrons is 1. Hint 1. and λ is the wavelength. 3. Calculate the photon energy Calculate the energy E of a photon with a wavelength of 290nm . Part A What is the maximum kinetic energy K0 of the photoelectrons when light of wavelength 290nm falls on the same surface? Use h = 6. Their mass is not well defined. Their speeds can only have certain quantized values. Energy of the photon Recall that the energy E of a photon is given by E = hf = hc λ . Their position can not be specified with absolute precision. Correct Light on a Photoelectric Surface When ultraviolet light with a wavelength of 400nm falls on a certain metal surface.All particles decay with a certain characteristic lifetime.10eV . Hint 1. Energy of the photon Recall that the energy E of a photon is given by E = hf = hc λ . 3. Work function definition The work function of a metal is the potential energy that a photoelectron must overcome before it can be emitted from the surface.28 eV Hint 2.28 eV Correct eV = 1. . and λ is the wavelength. Calculate the work function Calculate the work function ϕ for the metal surface in this problem.00×108m/s is the speed of light.E = 4. use the relation 1 ANSWER: K0 = 2. Hint 1. Express your answer in electron volts.60 × 10 −19 J . Hint 2.01 eV Hint 3. ANSWER: ϕ = 2.63×10−34J ⋅ s is Planck's constant. Converting to electron volts To convert your answers into electron volts. where 6. given the energy of the photon with a wavelength of 400nm and the emitted photoelectron with an energy of 1. It is essentially the binding energy of the electron to the metal.10eV . f is the frequency. 3 Lithium 2. Metal Φ (eV) Cesium 1.0 eV of kinetic energy.5 Calcium 3.5 Silver 4. Find the energy of the incident photon How much energy E does a 190nm photon have? Enter your answer numerically in electron volts to two significant figures. .Photoelectric Effect The following table lists the work functions of a few common metals.2 Copper 4.5 eV = hν .2 Sodium 2.6 Using these data. measured in electron volts.7 Platinum 5. answer the following questions about the photoelectric effect. Which of the metals in the table is the surface most likely to be made of? Hint 1. Energy of a photon The energy of a photon of frequency ν is E ANSWER: E = 6.9 Potassium 2. Frequency and wavelength are related as λν = c . The most energetic electrons emitted from the surface are measured to have 4. Hint 1. Part A Light with a wavelength of 190 nm is incident on a metal surface. Energy of a photon The energy of a photon of frequency ν is E ANSWER: = hν . ANSWER: cesium potassium sodium lithium calcium copper silver platinum Correct Part B Of the eight metals listed in the table. Frequency and wavelength are related as λν = c . . Hint 1. Photoelectric effect The work function is the energy required to remove an electron from the metal. Find the energy of the incident photon How much energy E does a 510nm photon have? Enter your answer numerically in electron volts to two significant figures. by conservation of energy. Thus. how many will eject electrons when a green laser (λ g = 510 nm ) is shined on them? Hint 1. the work function plus the kinetic energy of the emitted electron will be equal to the energy of the incident photon.Hint 2. If the work function is smaller. When a photon is absorbed by a metal. the metal will only emit an electron if the energy of the incident photon is greater than the work function of the metal. less energy is required to remove an electron. . How to approach the problem Recall that the work function represents the energy required to remove an electron from a metal.7 eV of kinetic energy.4 eV Hint 2. ANSWER: 0 1 2 3 4 5 6 7 8 Correct Part C Light with some unknown wavelength is incident on a piece of copper. Hint 1.E = 2. Photoelectric effect The work function is the energy required to remove an electron from the metal. If the copper is replaced with a piece of sodium. The most energetic electrons emitted from the copper have 2. what will be the maximum possible kinetic energy K of the electrons emitted from this new surface? Enter your answer numerically in electron volts to two significant figures. Correct The larger the electron's momentum. the de Broglie wavelength of an electron is inversely proportional to its momentum. the shorter its de Broglie wavelength will be. The de Broglie wavelength of the electron is unchanged.9 eV Correct The de Broglie Wavelength Part A How does the de Broglie wavelength of an electron change if its momentum increases? Hint 1. ANSWER: The de Broglie wavelength of the electron increases. where h is Planck's constant. .ANSWER: K = 4. The de Broglie wavelength The de Broglie wavelength λ of an electron of momentum p is defined as λ= h p . That is. The de Broglie wavelength of the electron decreases. the larger its de Broglie wavelength will be. Note that the de Broglie wavelength of an electron is inversely proportional to its speed.Part B How does the de Broglie wavelength of an electron change if its kinetic energy decreases? Hint 1. then.60×10−19C for the charge on an electron. Correct The lower the electron's energy. ± Accelerating Electrons Part A Through what potential difference ΔV must electrons be accelerated (from rest) so that they will have the same wavelength as an xray of wavelength 0. ANSWER: The de Broglie wavelength of the electron increases. The de Broglie wavelength of the electron is unchanged. where m and v are the mass and speed of the electron. p can be expressed as λ= h mv = mv . How to approach the question Recall that magnitude of an object's momentum is the product of its mass and speed. 9. . The de Broglie wavelength of the electron decreases. .63×10−34J ⋅ s for Planck's constant.135nm ? Use 6. Use this information along with the relation between kinetic energy and speed of an object to determine how the de Broglie wavelength of an electron changes with its kinetic energy. and 1.11×10−31kg for the mass of an electron. The de Broglie wavelength of an electron. This agrees with the result from Part A. Express your answer using three significant figures. How to approach the problem When an electron is accelerated through a potential difference. ANSWER: ΔV qe ΔV − q e qe ΔV −q e ΔV Hint 3. Kinetic energy as a function of momentum = 1 2 .11×10−31kg for the mass of an electron.Hint 1. Hint 2. Hint 1. Find the work done on an electron by the electric force An electron. Find the kinetic energy of the electrons What is the kinetic energy K of an electron that has the same wavelength as an xray of wavelength 0. Potential difference From the definition of electric potential as electric potential energy per unit charge. it follows that the potential difference VAB = VA − VB .63×10−34J ⋅ s for Planck's constant and 9. the difference between the potential at point A and the potential at point B.135nm ? Use 6. is accelerated to point b through a potential difference ΔV the electric force? = Vb − Va . you can write an expression that links the accelerating potential to the kinetic energy of the electron. Hint 1. initially at point a. the change in kinetic energy of the electron equals the work done on the electron by the electric force. Express your answer using three significant figures. What is the work done on the electron by Let q e be the magnitude of the charge of an electron. you need to express the kinetic energy of the electrons in terms of their wavelength. Thus. equals the work done by the electric force when a unit charge moves from A to B. Then. Express your answer using three significant figures.32×10−17 J ANSWER: ΔV = 82.63×10−34J ⋅ s for Planck's constant. ANSWER: p = 4.91×10−24 kg ⋅ m/s ANSWER: K = 1. the kinetic energy of the particle can also be expressed as K= p 2 2m 1 2 mv 2 . given the momentum p . Hint 1. = mv of the .63×10−34J ⋅ s is Planck's constant. However. Hint 2. Find the momentum of the electrons What is the momentum p of an electron that has the same wavelength as an xray of wavelength 0.135nm ? Use 6.The kinetic energy of a particle with mass m and velocity v is defined as K = particle. The de Broglie wavelength The de Broglie wavelength of a particle with momentum p is defined as λ= where h = 6.5 V Correct h p . 135nm ? Recall that xrays are electromagnetic waves and are subject to the same quantum relations as those of photons of light.00×108m/s for the speed of light in a vacuum. it is useful to write an expression that links the accelerating potential to the kinetic energy of the electrons. Hint 1. Find the energy of the xray What is the energy E of an xray of wavelength 0.00×108m/s is the speed of light and h = 6. How to approach the problem As in the previous part. its energy is E=h c λ .Part B Through what potential difference ΔV must electrons be accelerated so they will have the same energy as the xray in Part A? Use 6. Express your answer using three significant figures. Express your answer using three significant figures.60×10−19C for the charge on an electron.63×10−34J ⋅ s for Planck's constant.63×10−34J ⋅ s is Planck's constant. The energy of a photon Given the wavelength λ of a photon.47×10−15 J ANSWER: ΔV = 9190 V . Use 6.00×108m/s for the speed of light in a vacuum.63×10−34J ⋅ s for Planck's constant and 3. ANSWER: E = 1. this time you need to express the energy of the electrons in terms of the energy of xrays of given wavelength. and 1. Hint 1. 3. However. where c = 3. Hint 2. Both abovementioned effects occur. Neither first effect nor second one take place. the photoelectrons are faster.28 In a photoelectric effect experiment. the intensity of the light is increased while the frequency is held constant. what is the work function of the photocathode? Express your answer with the appropriate units. Part A As a result. ANSWER: there are more photoelectrons. Electrons emitted by the photoelectric effect are accelerated and then strike a phosphorescent screen. in the infrared: Part A If the threshold wavelength is 900 nm. causing it to glow more brightly than the original scene. ANSWER: . Recent devices are sensitive to wavelengths as long as 900 nm.Correct Multiple Choice Question 28. Correct Problem 28.14 Image intensifiers used in nightvision devices create a bright image from dim light by letting the light first fall on a photocathode. 14nm ? Express your answer using two significant figures.30 Part A What is the speed of an electron with a de Broglie wavelength of 0. in eV.E = 1.38 eV Correct Part B If light of wavelength 720nm strikes such a photocathode. ANSWER: K = 0.35 eV Correct Problem 28. ANSWER: v = 5. of the emitted electrons? Express your answer using two decimal places and include the appropriate units. what will be the maximum kinetic energy.2×106 m/s Correct Part B . and 9.0 eV .42 The allowed energies of a quantum system are 0. and 6.0 eV.0 eV. Part A What wavelengths appear in the system's emission spectrum? Enter your answers in ascending order separated by commas. ANSWER: v = 2800 m/s Correct Problem 28.0 eV.What is the speed of a proton with a de Broglie wavelength of 0. 1.5 eV.410 nm Correct Problem 28. 3. 6.44 The allowed energies of a quantum system are 0.0 eV . ANSWER: λi = 140. Part A How many different wavelengths appear in the emission spectrum? ANSWER: .210.0 eV.14nm ? Express your answer using two significant figures. The graph of the figure shows how the stopping potential depended on the frequency of the light. in eV.54 Light of constant intensity but varying wavelength was used to illuminate the cathode in a photoelectriceffect experiment. Part A What is the work function.N = 4 Correct Problem 28. ANSWER: E0 = 4 eV . of the cathode? Express your answer using one significant figure. 9×106m/s pass through a doubleslit apparatus.0mm ? Express your answer using two significant figures. Interference fringes are detected with a fringe spacing of 1. ANSWER: Δy n = 0.68 Electrons with a speed of 2.Correct Problem 28.8588 .0mm . Part A What will the fringe spacing be if the electrons are replaced by neutrons with the same speed? Express your answer using two significant figures. ANSWER: vn = 1600 m/s Correct Problem 28.55 μm Correct Part B What speed must neutrons have to produce interference fringes with a fringe spacing of 1. Part A What happens when the xray photon scatters from the electron? ANSWER: Its speed increases Its speed decreases Its speed stays the same Correct Part B . which recoils at a high speed. It has a collision with a slowmoving electron. and the photon energy formula E = hf tells us that its frequency must decrease. changing direction and frequency in the process.Further support for the photon model of electromagnetic waves comes from Compton scattering. but the photon model can. Suppose an xray photon is moving to the right. The collision looks very much like the collision between two particles. Classical electromagnetic wave theory cannot explain the change in frequency of the xrays on scattering. in which xrays scatter from electrons. The photon transfers energy and momentum to the electron. The xray photon loses energy. as in the figure. Which statement offers the best comparison between Compton scattering and xray diffraction? ANSWER: .What happens when the xray photon scatters from the electron? ANSWER: Its wavelength increases Its wavelength decreases Its wavelength stays the same Correct Part C What happens when the electron is struck by the xray photon? ANSWER: Its de Broglie wavelength increases Its de Broglie wavelength decreases Its de Broglie wavelength stays the same Correct Part D Xray diffraction can also change the direction of a beam of xrays. 58×105 m/s Correct Part B Calculate the work function.888V .15 Light with a wavelength of 390nm shines on a metal surface. which emits electrons. ANSWER: A = 2. The stopping potential is measured to be 0.Xray diffraction changes the wavelength of xrays; Compton scattering does not Compton scattering changes the speed of xrays; xray diffraction does not Xray diffraction relies on the particle nature of the xrays; Compton scattering relies on the wave nature Xray diffraction relies on the wave nature of the xrays; Compton scattering relies on the particle nature Correct Problem 28. Part A What is the maximum speed of emitted electrons? ANSWER: v = 5.30 eV Correct . ANSWER: Potassium Aluminum Iron Gold Correct Problem 28. Part A From what metal is the cathode made? ANSWER: gold aluminum silver potassium Correct Part B .Part C Identify the metal.93 V when light of 200 nm is used to illuminate the cathode.12 A photoelectriceffect experiment finds a stopping potential of 1. ANSWER: λ = 4100 nm Correct Part B Is this wavelength visible.20 Part A What is the wavelength. ultraviolet.93 V Correct Problem 28. of a photon with energy 0.30 eV? Express your answer using two significant figures. or infrared light? ANSWER: visible light ultraviolet light infrared light . in nm.What is the stopping potential if the intensity of the light is doubled? ANSWER: V = 1. of a photon with energy 30 eV? Express your answer using two significant figures. ultraviolet. ANSWER: λ = 410 nm Correct Part D Is this wavelength visible. or infrared light? ANSWER: visible light ultraviolet light infrared light Correct Part E What is the wavelength.Correct Part C What is the wavelength. of a photon with energy 3. in nm.0 eV? Express your answer using two significant figures. in nm. ANSWER: . . or infrared light? ANSWER: visible light ultraviolet light infrared light Correct Multiple Choice Question 28. Both abovementioned effects. Neither first effect nor second one.λ = 41 nm Correct Part F Is this wavelength visible.27 In a photoelectric effect experiment. ultraviolet. ANSWER: There are more electrons. Part A As a result. the frequency of the light is increased while the intensity is held constant. The electrons are faster. emitting photon R. The spacing between energy levels is drawn to scale. The electron jumps to n jumps to n = 1. = 2 . and then Part A The spacing between energy levels is drawn to scale.35 Photon P in the figure moves an electron from energy level n = 1 to energy level n = 3.Correct Multiple Choice Question 28. emitting photon Q. What is the correct relationship among the wavelengths of the photons? ANSWER: λP < λQ < λR λR < λP < λQ λQ < λP < λR λP < λR < λQ . 1890nm . 388nm . Part A What is the longest wavelength of electromagnetic radiation that the sensor can detect? ANSWER: 1390nm . Part A Why is this true? ANSWER: . electrons are never emitted from a metal if the frequency of the incoming light is below a certain threshold value.Correct Multiple Choice Question 28. Correct Multiple Choice Question 28. 888nm .26 A light sensor is based on a photodiode that requires a minimum photon energy of 1.29 In the photoelectric effect.40eV to create mobile electrons. After noting the properties of the pattern. The fringes would get farther apart.Photons of lowerfrequency light don't have enough energy to eject an electron. Part A What effect would this have? ANSWER: The fringes would get closer together. The positions of the fringes would not change. Correct Conceptual Question 28. The number of photons in lowfrequency light is too small to eject electrons. Lowfrequency light does not penetrate far enough into the metal to eject electrons.34 You shoot a beam of electrons through a double slit to make an interference pattern. . The electric field of lowfrequency light does not vibrate the electrons rapidly enough to eject them. you then double the speed of the electrons.5 shows the typical photoelectric behavior of a metal as the anodecathode potential difference ΔV is varied. Correct Multiple Choice Question 28. A positive anode attracts all of the electrons emitted by the cathode. the current incareases by some value and stays the same until the next large enough increase in ΔV . once all of the electrons reach the cathode. Since the number of the electrones emitted by the cathode is limited. A positive anode attracts all of the electrons emitted by the cathode.Part A Why do the curves become horizontal for ΔV > 1 V ? Shouldn't the current increase as the potential difference increases? ANSWER: A positive cathode attracts all of the electrons emitted by the anode. A positive cathode attracts all of the electrons emitted by the anode. Correct Part B Δ < 0 V Δ < 0 V . the current incareases by some value and stays the same until the next large enough increase in ΔV . a further increase in ΔV does not cause any further increase in the current. Once ΔV is large enough to make a banch of electrons leave the cathode. Since the number of the electrones emitted by the anode is limited. a further increase in ΔV does not cause any further increase in the current. Once ΔV is large enough to make a banch of electrons leave the anode. once all of the electrons reach the anode. Where do the electrons go? Are no electrons emitted or if they are. If an electron has enough potential energy when it leaves the cathode it might just reach the anode even when ΔV < 0 V . Correct Part C The current is zero for ΔV < −2.Why doesn't the current immediately drop to zero for ΔV < 0 V ? Shouldn't ΔV < 0 V prevent the electrons from reaching the anode? ANSWER: If an electron has enough kinetic energy when it leaves the cathode it might just reach the anode even when ΔV < 0 V .18 Part A Can an electron with a de Broglie wavelength of 2 μm pass through a slit that is 1 μm wide? Select the correct answer and explanation. Correct Conceptual Question 28. If an electron has enough potential energy when it leaves the anode it might just reach the cathode even when ΔV < 0 V . Electrons are not emmited for ΔV < −Vcritical . Electrons are not emmited for ΔV < −Vcritical . but they do not have sufficient kinetic energy to reach the cathode. ANSWER: . why is there no current? ANSWER: Electrons are still emitted. which is defined by the material of the cathode.0 V . If an electron has enough kinetic energy when it leaves the anode it might just reach the cathode even when ΔV < 0 V . which is defined by the material of the anod. but they do not have sufficient kinetic energy to reach the anode. Electrons are still emitted. which has a work function of 4.7 Part A What is the maximum kinetic energy of the ejected electrons? ANSWER: 1.0 eV photons is incident on a piece of iron. Yes. . It is impossible to say because electron can only pass through a slit that is wider than the half of the electron wavelength. No.0 eV eV Correct Reading Question 28. Electron can only pass through a slit that is wider than the electron wavelength. Electron wavelength is only a measure of how rapidly its amplitude varies as measured along its direction of motion and has nothing to do with passing through the slit. Correct Multiple Choice Question 28.Yes.32 Light consisting of 6.01 Part A eV . but electron can pass through the slit due to quantum tunneling. More precise values of the wavelength and the slit width are necessary.7 eV 10.7 6.3 eV 4. If the electron was a classic particle it would be impossible. ANSWER: electrons have a wave nature a photon can be converted into an electron electrons are the conductors in metals light has a particle nature Correct Reading Question 28. Hint 1.The photoelectric effect tells us that __________. ANSWER: its mass its charge its frequency its speed Correct .02 Part A The energy of a photon depends on __________. Correct Reading Question 28.Reading Question 28. ANSWER: emits an electron emits a photon emits a plasmon emits a neutron Correct .04 Part A When an electron in a quantum system drops from a higher energy level to a lower one.03 Part A Which of the following wave properties are exhibited by particles? ANSWER: diffraction superposition interference All of the listed answers are correct. the system __________. ANSWER: destroy information about its speed destroy its wave nature cause it to diffract cause the particle to be annihilated Correct Score Summary: Your score on this assignment is 110%. plus 3 points of extra credit.Reading Question 28. you __________.77 out of a possible total of 28 points. You received 27. .05 Part A If you precisely measure the position of a particle.