Inha University Department of Physics¡ ¼sol¡ ½ Whether in Cartesian (x, y , z ) or spherical coordinates, three quantities are needed to describe the variation of the wave function throughout space. The three quantum numbers needed to describe an atomic electron correspond to the variation in the radial direction, the variation in the azimuthal direction (the variation along the circumference of the classical orbit), and the variation with the polar direction (variation along the direction from the classical axis of rotation). Chapter 6 Problem Solutions 1. Why is it natural that three quantum numbers are needed to describe an atomic electron (apart from electron spin)? 3. Show that ¡ ¼sol¡ ½ For the given function, is a solution of Eq. (6.14) and that it is normalized. o a r e a r R 2 2 3 0 10 2 / / ) ( − · and 2 2 5 0 10 , / / o a r e a R dr d − − · Inha University Department of Physics 10 2 2 2 2 5 10 2 2 2 1 2 1 2 1 R a r a e a r r r a dr dR r dr d r o o a r o o o , _ ¸ ¸ − · , _ ¸ ¸ − − · , _ ¸ ¸ − / / This is the solution to Equation (6.14) if l=0 ( as indicated by the index of R 10 ), 2 2 2 2 2 4 or 4 2 2 me h a h me a o o o o ε π πε · · , which is the case, and 1 2 2 2 8 or 1 2 E a e E a E m o o o · − · − · πε , h again as indicated by the index of R 10 . To show normalization, , / ∫ ∫ ∫ ∞ − ∞ − ∞ · · 0 2 0 2 2 3 0 2 2 10 2 1 4 du e u dr e r a dr r R u a r o o where the substitution u=2r/a o has been made. The improper definite integral in u is known to have the value 2 and so the given function is normalized. Inha University Department of Physics 5. In Exercise 12 of Chap. 5 it was stated that an important property of the eigenfunctions of a system is that they are orthogonal to one another, which means that ¡ ¼sol¡ ½ From Equation (6.15) the integral, apart from the normalization constants, is m n dV m n ≠ · ∫ ∞ ∞ − 0 ψ ψ * l l m m m m d l l ′ ≠ Φ Φ ∫ ′ for 2 0 φ π * Verify that this is true for the azimuthal wave functions of the hydrogen atom by calculating l m Φ , * φ φ π φ φ π d e e d l l l l m i im m m ∫ ∫ ′ − ′ · Φ Φ 2 0 2 0 It is possible to express the integral in terms of real and imaginary parts, but it turns out to be more convenient to do the integral directly in terms of complex exponentials: [ ] 0 1 2 0 2 0 2 0 · − ′ · · ′ − ′ ′ − ′ ′ − ∫ ∫ π φ π φ π φ φ φ φ ) ( ) ( ) ( l l l l l l m m i l l m m i m i im e m m i d e d e e The above form for the integral is valid only for m l ≠ m l ’ , which is given for this case. In evaluating the integral at the limits, the fact that e i2πn = 1 for any integer n ( in this case (m l ’ – m l )) has been used. Inha University Department of Physics 7. Compare the angular momentum of a ground-state electron in the Bohr model of the hydrogen atom with its value in the quantum theory. 9. Under what circumstances, if any, is L z equal to L? ¡ ¼sol¡ ½ In the Bohr model, for the ground-state orbit of an electron in a hydrogen atom, λ = h/ mv = 2πr, and so L = pr = . In the quantum theory, zero-angular-momentum states (ψ spherically symmetric) are allowed, and L = 0 for a ground-state hydrogen atom. ¡ ¼sol¡ ½ From Equation (6.22), L z must be an integer multiple of ; for L to be equal to L z , the product l(1+1), from Equation(6.21), must be the square of some integer less than or equal to l. But, or any nonnegative l, with equality holding in the first relation only if l = 0. Therefore, l(l + 1) is the square of an integer only if l = 0, in which case L z = 0 and L = L z = 0. 2 2 1 1 ) ( ) ( + < + ≤ l l l l Inha University Department of Physics 11. What are the possible values of the magnetic quantum number m l of an atomic electron whose orbital quantum number is l = 4? 13. Find the percentage difference between L and the maximum value of L z for an atomic electron in p, d , and f states. ¡ ¼sol¡ ½ From Equation (6.22), the possible values for the magnetic quantum number m l are m l = 0, ±1, ±2, ±3, ±4, a total of nine possible values. ¡ ¼sol¡ ½ The fractional difference between L and the largest value of L z , is, for a given l, . ) ( ) ( max , 1 1 1 1 + − · + − + · − l l l l l l l L L L z % . % . % . 13 13 0 - 1 and 3 state, a For 18 18 0 - 1 and 2 state, a For 29 29 0 - 1 and 1 state, a For 4 3 3 2 2 1 · · · · · · · · · l f l d l p Inha University Department of Physics 15. In Sec. 6.7 it is stated that the most probable value of r for a 1s electron in a hydrogen atom is the Bohr radius a o . Verify this. ¡ ¼sol¡ ½ Using R 10 (r) from Table 6.1 in Equation (6.25), The most probable value of r is that for which P(r) is a maximum. Differentiating the above expression for P(r) with respect to r and setting the derivative equal to zero, . ) ( / o a r o e a r r P 2 3 2 4 − · o o a r o o a r a r r e a r r a r P dr d o , , ) ( / 0 and or 0 2 2 4 2 2 2 3 · · · , _ ¸ ¸ − · − for an extreme. At r = 0, P(r) = 0, and because P(r) is never negative, this must be a minimum. dp/ dr → 0 as r → ∞, and this also corresponds to a minimum. The only maximum of P(r) is at r = a o , which is the radius of the first Bohr orbit. Inha University Department of Physics 17. Find the most probable value of r for a 3d electron in a hydrogen atom. ¡ ¼sol¡ ½ Using R 20 (r) from Table 6.1 in Equation (6.25), and ignoring the leading constants (which would not affect the position of extremes), o a r e r r P 3 2 6 / ) ( − · The most probable value of r is that for which P(r) is a maximum. Differentiating the above expression for P(r) with respect to r and setting the derivative equal to zero, o o a r o a r a r r e a r r r P dr d o 9 0 and 3 2 6 or 0 3 2 6 6 5 3 2 6 5 , , ) ( / · · · , _ ¸ ¸ − · − for an extreme. At r = 0, P(r) = 0, and because P(r) is never negative, this must be a minimum. dP/d r → 0 as r → ∞, and this also corresponds to a minimum. The only maximum of P(r) is at r = 9a o , which is the radius of the third Bohr orbit. Inha University Department of Physics 19. How much more likely is the electron in a ground-state hydrogen atom to be at the distance a o from the nucleus than at the distance 2a o ? ¡ ¼sol¡ ½ For the ground state, n = 1, the wave function is independent of angle, as seen from the functions Φ(φ) and Θ(θ) in Table 6.1, where for n = 1, l = m l , = 0 (see Problem 6-14). The ratio of the probabilities is then the ratio of the product r 2 (R 10 (r)) 2 evaluated at the different distances. Specially, 47 1 4 4 1 2 2 1 2 2 2 2 2 2 . ) / ( ) / ( ) / ( ) ( / ) / ( / · · · · − − − − e e e e a e a dr a P dr a P o o o o a a o a a o o o 85 1 4 4 2 2 2 4 2 2 2 2 2 2 . ) ( ) ( ) ( ) ( / ) ( / · · · · − − − − e e e e a e a dr a P dr a P o o o o a a o a a o o o Similarly, Inha University Department of Physics 21. The probability of finding an atomic electron whose radial wave function is R(r) outside a sphere of radius r o centered on the nucleus is (a) Calculate the probability of finding a 1s electron in a hydrogen atom at a distance greater than a o from the nucleus. (b) When a 1s electron in a hydrogen atom is 2a o from the nucleus, all its energy is potential energy. According to classical physics, the electron therefore cannot ever exceed the distance 2a o from the nucleus. Find the probability r > 2a o for a 1s electron in a hydrogen atom. ¡ ¼sol¡ ½ (a) Using R 10 (r) for the 1s radial function from Table 6.1, ∫ ∞ o r dr r r R 2 2 ) ( , ) ( / du e u dr e r a dr r r R u a a r o a o o o ∫ ∫ ∫ ∞ − ∞ − ∞ · · 2 2 2 2 3 2 2 2 1 4 where the substitution u = 2r/a 0 has been made. Using the method outlined at the end of this chapter to find the improper definite integral leads to ( ) [ ] [ ] %, . 68 68 0 10 2 1 2 2 2 1 2 1 2 2 2 2 2 · · · + + − · − ∞ − ∞ − ∫ e u u e du e u u u Inha University Department of Physics (b) Repeating the above calculation with 2 a 0 as the lower limit of the integral, ( ) [ ] [ ] % , . 24 24 0 26 2 1 2 2 2 1 2 1 4 4 2 4 2 · · · + + − · − ∞ − ∞ − ∫ e u u e du e u u u 23. Unsold's theorem states that for any value of the orbital quantum number l, the probability densities summed over all possible states from m l = -1 to m l = +1 yield a constant independent of angles θ or φ that is, This theorem means that every closed subshell atom or ion (Sec. 7.6) has a spherically symmetric distribution of electric charge. Verify Unsold's theorem for l = 0, l = 1, and l = 2 with the help of Table 6. 1. ¡ ¼sol¡ ½For l = 0, only m l = 0 is allowed, Φ(φ) and Θ(θ) are both constants (from Table 6.1)), and the theorem is verified. For l = 1, the sum is constant 2 2 · Φ Θ ∑ + − · l l m l , s in cos s in π θ π θ π θ π 4 3 4 3 2 1 2 3 2 1 4 3 2 1 2 2 2 · + + Inha University Department of Physics . ) cos ( cos s in s in 2 2 2 2 4 1 3 16 10 2 1 4 15 2 1 2 16 15 2 1 2 − + + θ π θ θ π θ π The above may he simplified by extracting the commons constant factors, to ]. s in cos s in ) cos [( θ θ θ θ π 4 2 2 2 2 3 12 1 3 16 5 + + − Of the many ways of showing the term in brackets is indeed a constant, the one presented here, using a bit of hindsight, seems to be one of the more direct methods. Using the identity sin 2 θ = 1 - cos 2 θ to eliminate sin θ , , ) cos cos ( cos ) cos ( ) cos cos ( s in cos s in ) cos ( 1 2 1 3 1 12 1 6 9 3 12 1 3 4 2 2 2 2 4 4 2 2 2 2 · + − + − + + − · + + − θ θ θ θ θ θ θ θ θ θ and the theorem is verified. In the above, Φ*Φ= 1/2π, which holds for any l and m l , has been used. Note that one term appears twice, one for m l = -1 and once for m l = 1. For l = 2, combining the identical terms for m l = ±2 and m l = ±1, and again using Φ*Φ= 1/2π, the sum is Inha University Department of Physics 25. With the help of the wave functions listed in Table 6.1 verify that ∆l = ±1 for n = 2 àn = 1 transitions in the hydrogen atom. ¡ ¼sol¡ ½ In the integral of Equation (6.35), the radial integral will never. vanish, and only the angular functions Φ(φ) and Θ(θ) need to be considered. The ∆l = 0 transition is seen to be forbidden, in that the product π θ φ θ φ 4 1 00 0 00 0 · Θ Φ Θ Φ ∗ )) ( ) ( ( )) ( ) ( ( is spherically symmetric, and any integral of the form of Equation (6.35) must vanish, as the argument u = x, y or z will assume positive and negative values with equal probability amplitudes. If l = 1 in the initial state, the integral in Equation (6.35) will be seen to to vanish if u is chosen appropriately. If m l = 0 initially, and u = z = r cos θ is used, the integral (apart from constants) is 0 3 2 0 2 ≠ · ∫ θ θ θ π d s in cos If m l = ±1 initially, and u = x = r sin θ cos φ is used, the θ -integral is of the form 0 2 0 2 ≠ · ∫ π θ θ π d s in and the φ -integral is of the form 0 2 0 2 2 0 ≠ · · ∫ ∫ t π φ φ φ φ π π φ d d e i cos cos and the transition is allowed. Inha University Department of Physics 27. Verify that the n = 3 → n = 1 transition for the particle in a box of Sec. 5.8 is forbidden whereas the n = 3 → n = 2 and n = 2 → n = 1 transitions are allowed. ¡ ¼sol¡ ½ The relevant integrals are of the form . s in s in dx L x m L x n x L π π ∫ 0 The integrals may be found in a number of ways, including consulting tables or using symbolic- manipulation programs (see; for instance, the solution to Problem 5-15 for sample Maple commands that are easily adapted to this problem). One way to find a general form for the integral is to use the identity )] cos( ) [cos ( s in s in β α β α β α + − − · 2 1 and the indefinite integral (found from integration by parts) 2 1 k k x k k x x dx k x k k k x x dx k x x cos s in s in s in cos + · − · ∫ ∫ to find the above definite integral as , ) ( cos ) ( ) ( s in ) ( ) ( cos ) ( ) ( s in ) ( L L x m n m n L L x m n m n Lx L x m n m n L L x m n m n Lx 0 2 2 2 2 2 2 2 1 1 1 1 1 1 ] 1 ¸ + + + + + − − − + − − π π π π π π π Inha University Department of Physics where n ≠ m 2 is assumed. The terms involving sines vanish, with the result of . ) ( ) cos ( ) ( ) cos ( 1 ] 1 ¸ + − + − − − − 2 2 2 2 1 1 2 m n m n m n m n L π π π If n and m axe both odd or both even, n + m and n - m are even, the arguments of the cosine terms in the above expression are even-integral multiples of π, and the integral vanishes. Thus, the n = 3 → n = 1 transition is forbidden, while the n = 3 → n = 2 and n = 2 → n = 1 transitions are allowed. To make use of symmetry arguments, consider that dx L x m L x n x dx L x m L x n L x L L ∫ ∫ · , _ ¸ ¸ − 0 0 2 π π π π s in s in s in s in for n ≠ m, because the integral of L times the product of the wave functions is zero; the wave functions were shown to be orthogonal in Chapter 5 (again, see Problem 5-15). Letting u=L/ 2 – x, , _ ¸ ¸ − · − · L u n n L u L n L x n π π π π 2 2 s in ) ) / (( s in s in This expression will be ± cos ( nπu/L ) for n odd and ±sin ( nπu/L ) for n even. The integrand is then an odd function of u when n and m are both even or both odd, and hence the integral is zero. If one of n or m is even and the other odd, the integrand is an even function of u and the integral is nonzero. Inha University Department of Physics 29. Show that the magnetic moment of an electron in a Bohr orbit of radius r n is proportional to 31. Find the minimum magnetic field needed for the Zeeman effect to be observed in a spectral line of 400-nm wavelength when a spectrometer whose resolution is 0.010 nm is used. ¡ ¼sol¡ ½ From Equation (6.39), the magnitude of the magnetic moment of an electron in a Bohr orbit is proportional to the magnitude of the angular momentum, and hence proportional to n. The orbital radius is proportional to n 2 (See Equation (4.13) or Problem 4-28), and so the magnetic moment is proportional to . ¡ ¼sol¡ ½ See Example 6.4; solving for B, n r n r T 34 1 C) 10 6 1 m/s) 10 0 3 kg 10 1 9 4 m) 10 (400 m 10 010 0 4 19 8 31 2 9 - 9 2 . . ( . )( . ( . · × × × × × · ∆ · − − − π π λ λ e mc B 14) if l=0 ( as indicated by the index of R10). 2m 1 e2 E =− = E1 8πεoao or ao = 4π 2εoh 2 me 2 Inha University Department of Physics . ∫0 R10 r dr = a 3 ∫0 r e 2 o where the substitution u=2r/ao has been made. The improper definite integral in u is known to have the value 2 and so the given function is normalized. or h2 ao again as indicated by the index of R10. 4 ∞ 2 − 2r /ao 1 ∞ ∞ 2 2 dr = ∫0 u 2e −udu.1 d 2 dR10 2 1 r 2 −r /ao e r = − 5 /2 2 2r − 2 dr dr ao r ao r 1 2 R10 = 2− a r ao o This is the solution to Equation (6. To show normalization. and E = − 2 . 2 2me 2 = . ao h 2 4πεo which is the case. 5.15) the integral. but it turns out to be more convenient to do the integral directly in terms of complex exponentials: ∫0 2π e −imlφe imlφdφ = = ′ ∫0 2π i (ml′ −ml′ )φ e dφ 1 ′ ′ 2π e i(ml −ml )φ 0 = 0 i (ml′ − ml ) [ ] The above form for the integral is valid only for ml ≠ ml’. Inha University Department of Physics . 5 it was stated that an important property of the eigenfunctions of a system is that they are orthogonal to one another. In Exercise 12 of Chap. In evaluating the integral at the limits. apart from the normalization constants. the fact that ei2πn = 1 for any integer n ( in this case (ml’ – ml)) has been used. It is possible to express the integral in terms of real and imaginary parts. which is given for this case. which means that Verify that this is true for the azimuthal wave functions Φml of the hydrogen atom by calculating 2π * ∫ Φml Φml′ dφ for ml ≠ ml′ 0 ∞ * ∫−∞ψ nψm dV = 0 n ≠ m 【sol】 From Equation (6. is 2π * 2π −im φ im ′φ ∫0 Φml Φml′dφ = ∫0 e l e l dφ . with equality holding in the first relation only if l = 0. l(l + 1) is the square of an integer only if l = 0. in which case Lz = 0 and L = Lz = 0. Therefore. λ = h/mv = 2πr. l 2 ≤ l (l + 1) < (l + 1)2 or any nonnegative l. and so L = pr = . and L = 0 for a ground-state hydrogen atom. 【sol】 In the Bohr model. Compare the angular momentum of a ground-state electron in the Bohr model of the hydrogen atom with its value in the quantum theory. must be the square of some integer less than or equal to l. the product l(1+1). for the ground-state orbit of an electron in a hydrogen atom.7. Lz must be an integer multiple of . Inha University Department of Physics . zero-angular-momentum states (ψ spherically symmetric) are allowed.21). But. 9. In the quantum theory.22). if any. Under what circumstances. for L to be equal to Lz. is Lz equal to L? 【sol】 From Equation (6. from Equation(6. 【sol】 The fractional difference between L and the largest value of Lz. d. ±2. l = 2 and 1 For a f state. ±3. What are the possible values of the magnetic quantum number ml of an atomic electron whose orbital quantum number is l = 4? 【sol】 From Equation (6. 13.11. l (l + 1) l +1 = 0. and f states.max L For a p state.29 = 29% = 0. l = 3 and 1 1 2 2 3 3 4 = l (l + 1) − l l =1− . l = 1 and 1 For a d state. the possible values for the magnetic quantum number ml are ml = 0. for a given l. ±1.18 = 18% = 0. a total of nine possible values. ±4.22). is.13 = 13% Inha University Department of Physics . L − L z . Find the percentage difference between L and the maximum value of Lz for an atomic electron in p. Inha University Department of Physics . which is the radius of the first Bohr orbit. ao The most probable value of r is that for which P(r) is a maximum. Differentiating the above expression for P(r) with respect to r and setting the derivative equal to zero.7 it is stated that the most probable value of r for a 1s electron in a hydrogen atom is the Bohr radius ao. 6. 4r 2 − 2r /ao P (r ) = 3 e . ao or for an extreme.25). The only maximum of P(r) is at r = ao. and because P(r) is never negative. dr ao ao r2 r = ao and r = 0. dp/dr → 0 as r → ∞.15. Verify this. d 4 2r 2 − 2r /ao e P (r ) = 3 2r − = 0. P(r) = 0. and this also corresponds to a minimum. 【sol】 Using R10(r) from Table 6. At r = 0.1 in Equation (6. this must be a minimum. In Sec. At r = 0. which is the radius of the third Bohr orbit. P (r ) = r 6e − 2r / 3ao The most probable value of r is that for which P(r) is a maximum. 【sol】 Using R20 (r) from Table 6.17. P(r) = 0. and ignoring the leading constants (which would not affect the position of extremes). Inha University Department of Physics . Differentiating the above expression for P(r) with respect to r and setting the derivative equal to zero.1 in Equation (6. 9a o or for an extreme.25). dP/dr → 0 as r → ∞. and because P(r) is never negative. Find the most probable value of r for a 3d electron in a hydrogen atom. and this also corresponds to a minimum. The only maximum of P(r) is at r = 9ao. 3ao and r = 0. d P (r ) = dr 2r 6 6r = 3ao 5 5 2r 6 − 2r /3ao 6r − e = 0. this must be a minimum. l = ml. the wave function is independent of angle.19. Specially. n = 1. as seen from the functions Φ(φ) and Θ(θ) in Table 6. 2 P (ao )dr aoe o o e −2 e2 = = = = 1. 2 P (ao )dr a oe o o e −2 4 = = = = 1.1.85 P ( 2ao )dr ( 2a o )2 e − 2( 2ao )/ao ( 4)e − 4 4 − 2a /a Inha University Department of Physics . How much more likely is the electron in a ground-state hydrogen atom to be at the distance ao from the nucleus than at the distance 2ao? 【sol】 For the ground state. The ratio of the probabilities is then the ratio of the product r2 (R10 (r))2 evaluated at the different distances. where for n = 1.47 P (ao / 2)dr (ao / 2)2 e − 2(ao / 2) /ao (1/ 4)e −1 e − 2a /a Similarly. = 0 (see Problem 6-14). 21. Find the probability r > 2ao for a 1s electron in a hydrogen atom. 2 [ ] Inha University Department of Physics . (b) When a 1s electron in a hydrogen atom is 2ao from the nucleus. the electron therefore cannot ever exceed the distance 2ao from the nucleus. ∫ao R(r ) r dr = a 3 ∫ao r e 2 o where the substitution u = 2r/a0 has been made. The probability of finding an atomic electron whose radial wave function is R(r) outside a sphere of radius ro centered on the nucleus is ∫ ∞ ro R (r ) r 2dr 2 (a) Calculate the probability of finding a 1s electron in a hydrogen atom at a distance greater than ao from the nucleus.68 = 68%. all its energy is potential energy. Using the method outlined at the end of this chapter to find the improper definite integral leads to 1 ∞ 2 −u 1 −u 2 ∫2 u e du = 2 − e u + 2u + 2 2 [ ( )] ∞ 2 = 1 −2 e 10 = 0.1. 【sol】 (a) Using R10(r) for the 1s radial function from Table 6. ∞ 4 ∞ 2 − 2r /ao 1 ∞ 2 2 dr = ∫2 u 2e −udu . According to classical physics. Unsold's theorem states that for any value of the orbital quantum number l. ∞ 1 ∞ 2 −u 1 1 u e du = − e −u u 2 + 2u + 2 4 = e − 4 26 = 0. 7. For l = 1. ml = −l ∑Θ +l 2 Φ = constant 2 This theorem means that every closed subshell atom or ion (Sec. the probability densities summed over all possible states from ml = -1 to ml = +1 yield a constant independent of angles θ or φ that is. 2 ∫4 2 2 [ ( )] [ ] 23. 2π 4 2π 2 2π 4 4π Inha University Department of Physics . 1. 【sol】For l = 0. Φ(φ) and Θ(θ) are both constants (from Table 6.1)).6) has a spherically symmetric distribution of electric charge.(b) Repeating the above calculation with 2 a0 as the lower limit of the integral. and l = 2 with the help of Table 6. l = 1. only ml = 0 is allowed. the sum is 1 3 1 3 1 3 3 sin 2 θ + cos 2 θ + sin 2 θ = .24 = 24% . Verify Unsold's theorem for l = 0. and the theorem is verified. ( 3 cos 2 θ − 1)2 + 12 sin 2 θ cos2 θ + 3 sin 4 θ = ( 9 cos4 θ − 6 cos 2 θ + 1) + 12(1 − cos 2 θ ) cos 2 θ + 3(1 − 2 cos 2 θ + cos4 θ ) = 1.cos2 θ to eliminate sin θ . the sum is 2 1 15 1 15 1 10 sin4 θ + 2 sin2 θ cos2 θ + (3 cos2 θ − 1)2 . using a bit of hindsight. Inha University Department of Physics . one for ml = -1 and once for ml = 1. For l = 2. Using the identity sin2 θ = 1 . has been used. 2π 16 2π 4 2π 16 The above may he simplified by extracting the commons constant factors. and again using Φ*Φ= 1/2π. the one presented here. seems to be one of the more direct methods. Φ*Φ= 1/2π. and the theorem is verified. combining the identical terms for ml = ±2 and ml = ±1.In the above. to 5 [(3 cos2 θ − 1)2 + 12 sin2 θ cos2 θ + 3 sin4 θ ]. which holds for any l and ml. 16π Of the many ways of showing the term in brackets is indeed a constant. Note that one term appears twice. The ∆l = 0 transition is seen to be forbidden. and u = z = r cos θ is used. Inha University Department of Physics .1 verify that ∆l = ±1 for n = 2 à n = 1 transitions in the hydrogen atom. 【sol】 In the integral of Equation (6. in that the product 1 ( Φ0 (φ )Θ00(θ ))∗ (Φ 0 (φ )Θ00 (θ )) = 4π is spherically symmetric. If l = 1 in the initial state. the radial integral will never.25. and only the angular functions Φ(φ) and Θ(θ) need to be considered. and u = x = r sin θ cos φ is used.35). y or z will assume positive and negative values with equal probability amplitudes. the integral (apart from constants) is π 2 cos 2 θ sin θdθ = ≠ 0 ∫0 3 If ml = ±1 initially. the integral in Equation (6. the θ -integral is of the form and the φ -integral is of the form π 2 ∫0 sin θdθ = 2 ≠ 0 π ∫ 2π ±i φ e cos φdφ 0 = ∫ 2π cos2 φdφ 0 =π ≠0 and the transition is allowed. and any integral of the form of Equation (6. With the help of the wave functions listed in Table 6. If ml = 0 initially.35) will be seen to to vanish if u is chosen appropriately. as the argument u = x. vanish.35) must vanish. the solution to Problem 5-15 for sample Maple commands that are easily adapted to this problem).27. 2 Lx (n + m )πx L2 (n + m )πx − sin + cos 2 2 (n + m )π L L (n + m ) π 0 Inha University Department of Physics . ∫0 L L The integrals may be found in a number of ways. for instance.8 is forbidden whereas the n = 3 → n = 2 and n = 2 → n = 1 transitions are allowed. One way to find a general form for the integral is to use the identity sin α sin β = 1 [cos(α − β ) − cos(α + β )] 2 and the indefinite integral (found from integration by parts) x sin kx 1 x sin kx cos kx ∫ x cos kxdx = k − k ∫ sin kxdx = k + k 2 to find the above definite integral as L Lx (n − m )πx L2 (n − m )πx sin + cos (n − m ) L L (n − m )2 π 2 1 . 5. 【sol】 The relevant integrals are of the form nπx mπx L x sin sin dx. including consulting tables or using symbolicmanipulation programs (see. Verify that the n = 3 → n = 1 transition for the particle in a box of Sec. because the integral of L times the product of the wave functions is zero. 2π 2 (n − m )2 (n + m )2 If n and m axe both odd or both even. see Problem 5-15).where n ≠ m2 is assumed.m are even. To make use of symmetry arguments. The terms involving sines vanish. while the n = 3 → n = 2 and n = 2 → n = 1 transitions are allowed. the wave functions were shown to be orthogonal in Chapter 5 (again. the arguments of the cosine terms in the above expression are even-integral multiples of π. If one of n or m is even and the other odd. and the integral vanishes. the n = 3 → n = 1 transition is forbidden. with the result of L2 cos(n − m )π − 1 cos(n + m )π − 1 − . Inha University Department of Physics . n + m and n . Letting u=L/2 – x. Thus. and hence the integral is zero. sin nπx nπ (( L / 2) − u ) nπ nπu = sin = sin − L L 2 L This expression will be ± cos ( nπu/L ) for n odd and ±sin ( nπu/L ) for n even. The integrand is then an odd function of u when n and m are both even or both odd. consider that L L nπx mπx L nπx mπx sin dx = ∫0 x sin sin dx x − sin ∫0 2 L L L L for n ≠ m. the integrand is an even function of u and the integral is nonzero. 010 nm is used. solving for B.0 × 108 m/s) B = 2 = = 1. Find the minimum magnetic field needed for the Zeeman effect to be observed in a spectral line of 400-nm wavelength when a spectrometer whose resolution is 0.34 T -9 2 −19 e λ (400 × 10 m) (1. The orbital radius is proportional to n2 (See Equation (4. the magnitude of the magnetic moment of an electron in a Bohr orbit is proportional to the magnitude of the angular momentum. 31. and hence proportional to n.010 × 10−9 m 4π ( 9. and so the magnetic moment is proportional to rn . Show that the magnetic moment of an electron in a Bohr orbit of radius rn is proportional to rn 【sol】 From Equation (6.13) or Problem 4-28).4. ∆ λ 4πmc 0. 【sol】 See Example 6.29.39).6 × 10 C) Inha University Department of Physics .1 × 10−31 kg )( 3.