Physics exercises_solution: Chapter 40

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Physics exercises_solution: Chapter 40

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Bộ tài liệu bài tập tham khảo môn vật lý bậc đại học bằng tiếng anh Chapter 40

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Nội dung Text: Physics exercises_solution: Chapter 40

  1. n2h2 h2 (6.63  10 34 J  s) 2 40.1: a) En   E1   8mL2 8mL2 8(0.20 kg)(1.5 m) 2  E1  1.22  10 67 J. 1 2 2E 2(1.2  10 67 J) b) E  mv  v    1.1  10 33 m s 2 m 0.20 kg d 1.5 m t   33  1.4  1033 s. v 1.1  10 m s h2 3h 2 c) E2  E1  2 (4  1)  2  3(1.22  10 67 J)  3.7  10 67 J. 8mL 8mL d) No. The spacing between energy levels is so small that the energy appears continuous and the balls particle-like (as opposed to wave-like). h 40.2: L 8mE1 (6.626  10 34 J  s)   6.4  10 15 m.  27 19 8(1.673  10 kg)(5.0  10 eV)(1.602  10 6 J eV) h2 3h 2 3 40.3: E2  E1  (4  1)  Lh 8mL 2 8mL 2 8m( E2  E1 ) 3  (6.63  10 34 J  s) 31 8(9.11  10 kg)(3.0 eV)(1.60  10 19 J eV)  L  6.1  10 10 m  0.61 nm. 40.4: a) The energy of the given photon is c (3.00  103 m/s) E  hf  h  (6.63  10 34 J  s) 9  1.63  10 18 J. λ (122  10 m) The energy levels of a particle in a box are given by Eq. 40.9 h2 h 2 (n 2  m 2 ) (6.63  10 34 J  s) 2 (2 2  12 ) E  (n 2  m 2 )  L   8mL2 8mE 8(9.11  10 31 kg)(1.63  10 20 J)  3.33  10 10 m. b) The ground state energy for an electron in a box of the calculated dimensions is 6 63 1034 J 2 E  8 mL2  8(9.11(10.31kg)(3.33s )1010m) 2  5.43  10 19 J  3.40 eV (one-third of the original h2  photon energy), which does not correspond to the  13.6 eV ground state energy of the hydrogen atom. Note that the energy levels for a particle in a box are proportional to n 2 , whereas the energy levels for the hydrogen atom are proportional to  n12 .
  2. 40.5: E  1 mv 2  1 (5.00  10 3 kg)(0.010 m s) 2  2.5  10 7 J 2 2 h2 h b) E1  2 so L  8mL 8mE1 E1  2.5  10 7 J gives L  6.6  10 30 m 40.6: a) The wave function for n  1 vanishes only at x  0 and x  L in the range 0  x  L. b) In the range for x, the sine term is a maximum only at the middle of the box, x  L / 2. c) The answers to parts (a) and (b) are consistent with the figure. 2  2πx  40.7: The first excited state or (n  2) wave function is ψ 2 ( x)  sin . L  L  2 2  2πx  So ψ 2 ( x)  sin 2  . L  L   2πx  a) If the probability amplitude is zero, then sin 2  0  L  2πx Lm   mπ  x  , m  0, 1, 2 . . . L 2 L So probability is zero for x  0, , L. 2  2πx  b) The probability is largest if sin    1.  L  2πx π L   (2m  1)  x  (2m  1) . L 2 4 L 3L So probability is largest for x  and . 4 4 c) These answers are consistent with the zeros and maxima of Fig. 40.5. 40.8: a) The third excited state is n  4, so h2 E  (4 2  1) 8mL2 15(6.626  10 34 J  s) 2  31 9  5.78  10 17 J  361 eV. 8(9.11  10 kg)(0.125  10 m) 2 hc (6.63  10 34 J  s)(3.0  108 m/s) b) λ   E 5.78  10 17 J λ  3.44 nm.
  3. h h 40.9: Recall λ   . p 2mE h2 h a) E1   λ1   2 L  2(3.0  10 10 m)  6.0  10 10 m 8mL2 2 2mh / 8mL 2 34 h (6.63  10 J  s) p1    1.1  10 24 kg  m/s λ1 6.0  10 10 m 4h 2 b) E2   λ 2  L  3.0  10 10 m 8mL2 h p2   2 p1  2.2  10 24 kg  m/s. λ2 9h 2 2 c) E3  2  λ 3  L  2.0  10 10 m 8mL 3 24 p3  3 p1  3.3  10 kg  m/s. d 2ψ 8π 2 m 2m 40.10:  k ψ , and for ψ to be a solution of Eq. (40.3), k  E 2  E 2 . 2 2 dx 2 h  b) The wave function must vanish at the rigid walls; the given function will vanish at x  0 for any k , but to vanish at x  L, kL  nπ for integer n. h 2 d 2ψ 40.11: a) Eq.(40.3) :  .  Eψ. 8π 2 m dx 2 d2 d2 d 2 ψ  2 ( A cos kx)  ( Ak sin kx)   Ak 2 cos kx dx dx dx 2 2 Ak h  cos kx  EA cos kx 8π 2 m k 2h2 8π 2 mE 2mE E 2 k   . 8π m h 2  b) This is not an acceptable wave function for a box with rigid walls since we need ψ (0)  ψ ( L)  0, but this ψ (x) has maxima there. It doesn’t satisfy the boundary condition. 2 d 2ψ  2 d 2ψ 40.12:   Uψ    C  UCψ 2m dx 2 2m dx 2  2 d 2ψ   C  2  Uψ   2m dx   CEψ  ECψ  Eψ , and so ψ  is a solution to Eq. (40.1) with the same energy.
  4.  2 d 2ψ 40.13: a) Eq.(40.1) :  Uψ  Eψ 2m dx 2  2 d 2 Left-hand side: ( A sin kx)  U 0 A sin kx 2m dx 2 2 k 2  A sin kx  U 0 A sin kx 2m  2 k 2    2m  U 0 ψ.    2 k 2 2 k 2 But  U 0  U 0  E for constant k . But  U 0 should equal E  no solution. 2m 2m 2 k 2 b) If E  U 0 , then  U 0  E is consistent and so ψ  A sin kx is a solution of 2m Eq.(40.1) for this case. 40.14: According to Eq.40.17, the wavelength of the electron inside of the square well is given by 2mE h k  λ in  .  2m(3U 0 ) By an analysis similar to that used to derive Eq.40.17, we can show that outside the box h h λ out   . 2 m( E  U 0 ) 2m(2U 0 ) Thus, the ratio of the wavelengths is λ out 2m(3U 0 ) 3   . λ in 2m(2U 0 ) 2 π 2 2 40.15: E1  0.625E  0.625 ; E1  2.00 eV  3.20  10 19 J 2mL2 1/ 2  0.625  L  π 2(9.109  10 31 kg)(3.20  10 19 J)   3.43  10 m  10   40.16: Since U 0  6 E we can use the result E1  0.625 E from Section 40.3, so U 0  E1  5.375 E and the maximum wavelength of the photon would be hc hc 8mL2 c λ   U 0  E1 (5.375)(h 2 8mL2 ) (5.375)h 8(9.11  10 31 kg)(1.50  10 9 m) 2 (3.00  108 m/s)  34  1.38  10 6 m. (5.375)(6.63  10 J  s)
  5. 40.17: Since U 0  6 E  , we can use the results from Section 40.3, E1  0.625E , E3  5.09 and π 2 2 π 2 (1.054  10 34 J  s.) 2 E   2mL2 2(1.67  10  27 kg)(4.0  10 15 m) 2  E  2.05  10 12 J. The transition energy is E3  E1  (5.09  0.625)(2.05  10 12 J)  9.15  10 12 J. The wavelength of the photon absorbed is then hc (6.63  10 34 J  s)(3.00  108 m/s) λ  12  2.17  10 14 m. E3  E1 9.15  10 J 40.18: Since U 0  6 E , we can use the results from Section 40.3. h2 E  . 8mL2 E2  2.43E and E1  0.625E . hc 1.805h 2 E  E2  E1   (2.43  0.625) E  . λ 8mL2 (1.805)hλ (1.805)(6.63  10 34 J  s)(4.55  10 7 m) So L   31  4.99  10 10 m 8mc 8(9.11  10 kg)(3.00  10 m s) 8 2mE 2mE 40.19: Eq.(40.16) : ψ  Asin x  B cos x   d 2ψ  2mE  2mE  2mE  2mE   A 2  sin x  B 2  cos x         2 dx  2mE  (ψ )  Eq.(40.15). 2 dψ 40.20:   (Cex  De x ), dx d 2ψ 2   2 (Cex  De x )   2ψ dx for all constants C and D. Hence ψ is a solution to Eq. (40.1) for 2 2    U 0  E , or   [2m(U 0  E )]1/ 2 , 2m and  is real for E  U 0 .
  6. E  E  2 2 m (U 0  E ) 40.21: T  16 1   U e   L. U0 0  E 6.0 eV  and E  U 0  5 eV  8.0  10 19 J. U 0 11.0 eV a) L  0.80  10 9 m  6.0 eV   6.0 ev  2 2 ( 9.1110 31 kg)(8.010 19 J) T  16  1   11.0 eV   11.0 eV e  1.05510  34 Js ( 0.80  10 9 m)    8  4.4  10 . b) L  0.40  10 9 m T  4.2  10 4. 40.22: (Also see Problem 40.25). The transmission coefficient is E  E  2 2 m (U 0  E ) L T  16 1   U e   with E  5.0 eV, L  0.60  10 9 m, and U0  0  m  9.11  10 31 kg a) U 0  7.0 eV  T  5.5  104. b) U 0  9.0 eV  T  1.8  10 5 c) U 0  13.0 eV  T  1.1  10 7. 40.23: λ  h p  h 2mK , so λ K is constant λ1 K1  λ 2 K 2 ; λ1 and K1 are for x  L where K1  2U 0 and λ 2 and K 2 are for 0  x  L where K 2  E  U 0  U 0 λ1 K2 U0 1    λ2 K1 2U 0 2 40.24: Using Eq. 40.21 E  E  12.0 eV  12.0 eV  G  16 1   U   16 15.0 eV 1  15.0 eV   2.56    U0  0    2m(U 0  E ) 2(9.11  10 31 kg)(15.0  12.0 eV)(1.60  10 19 J/eV)     (6.63  10 34 J  s) 2π 8.9  109 m 1 1 1  2.56  T  Ge 2L  L  ln(G T )  ln   0.26 nm. 2 1 2(8.9  10 m )  0.025  8
  7. 40.25: a) Probability of tunneling is T  Ge 2L E  E   32   32  where G  16 1   U   16  41  1  41   2.74  U0  0     2m(U 0  E ) 2(9.11  10 31 kg)(41eV  32 eV)(1.60  10 9 J/eV). and     1.054  10 34 J  s  1.54  1010 m 1. 10 m 1 )( 2.51010 m) So Te  2.74 e  2(1.54  10  2.74e 7.7  1.2  10 3. mp 1.67  10 27 b) For a proton,     . me 9.11  10 31     6.59  1011 m 1 11 m 1 )( 2.5  10 10 m)  T  2.74e 2( 6.59  10  2.74e 330  T  10 143. The is small and for all practical purposes equal to zero. E  E  2m(U 0  E ) 40.26: T  Ge 2L with G  16 1   U and    , U0  0   2 2 m (U 0  E ) E  E  L Giving T  16 1   U e   . U0  0  a) If U 0  30.0  10 6 eV, L  2.0  10 15 m, m  6.64  10 27 kg and U 0  E  1.0  106 eV( E  29.0  106 eV), T  0.090. b) If U 0  E  10.0  10 6 eV ( E  20.0  10 6 eV) T  0.014. 40.27: The ground state energy of a simple harmonic oscillator is, with n  0, 1 1 k  (1.055  10 34 J  s) 110 N/m E0  ω    2 2 m 2 0.250 kg  E0  1.11  10 33 J  6.91  10 15 eV Also En1  En  ω  2 E0  2.22  10 33 J  1.38  10 14 eV. Such tiny energies are unimportant for the motion of the block so quantum effects are not important. dψ d 2ψ 40.28: Let mk  2  δ , and so  2 xψ and 2  (4 x 2 δ 2  2δ)ψ , and ψis a dx dx 2 1 1 solution of Eq.(40.21) if E  δ   k /m  ω. m 2 2
  8. hc 40.29: The photon’s energy is E  . λ 3 1 k The transition energy is E  E1  E0  ω     ω    2 2 m 2πc k 4π 2 c 2 m 4π 2 (3.00  108 m s) 2 (9.4  10 26 kg)    k   λ m λ2 (5.25  10 4 m) 2  k   1.2 N m . 40.30: According to Eq.40.26, the energy released during the transition between two adjacent levels is twice the ground state energy E3  E2  ω  2 E0  11.2 eV. For a photon of energy E c hc (6.63  10 34 J. s)(3.00  108 m s) E  hf  λ     111 nm. f E (11.2 eV)(1.60  10 19 J/eV) 40.31: a) E  (n  1 ) ω 2 For n  1, E  ω   k m  1.544  10 21 J E  hc λ so λ  (hc) E  129 μm b) E0  1 ω  1 (1.544  10 21 J)  4.82 meV 2 2 2 ψ ( A)  mk  2   ω 40.32: a)  exp   A   exp  mk    e 1  0.368.  ψ (0) 2     k  This is more or less what is shown in Fig. (40.19). ψ (2 A) 2  mk    ω b)  exp    (2 A) 2   exp   mk  4   e 4  1.83  10 2.  ψ (0) 2     k  This figure cannot be read this precisely, but the qualitative decrease in amplitude with distance is clear.
  9. 40.33: For an excited level of the harmonic oscillator  1 1 En   n   ω  k A2  2 2 (2n  1)ω  A . This is the uncertainty in the position. k  1 1 Also En   n   ω  mv 2 max  2 2 (2n  1)ω  vmax  m (2n  1)ω  xp  A  (mvmax )   (2n  1)ωm k m  (2n  1)ω  (2n  1). k So xp  (2n  1), which agrees for the ground state (n  0) with xp  . The uncertainty is seen to increase with n. 2 L/4 πx 2 L/4 1  2πx  40.34: a) L 0 sin 2 dx   L L 0 2 1  cos L   dx L 4 1 L 2πx   x  sin  L 2 L 0 1 1   , 4 2π about 0.0908. b) Repeating with limits of L 4 and L 2 gives L2 1 L 2πx  1 1 x  sin    , L 2π L  L 4 4 2 about 0.0409. c) The particle is much likely to be nearer the middle of the box than the edge. d) The results sum to exactly 1 2 , which means that the particle is as likely to be between x  0 and L 2 as it is to be between x  L 2 and x  L. e) These results are represented in Fig. (40.5b).
  10. 3L / 4 3L / 4 2  πx  40.35: a) P   | ψ1 |2 dx   sin 2   dx L/4 L/4 L L πx π Let z   dz  dx L L 3π 4 2 3 / 4 1 1  P  sin z d z   z  sin 2 z  2  /4 π 2 π 4 1 1    0.818. 2 π 3L / 4 3L / 4 2 2  2πx  b) P   | ψ 2 |2 dx   sin   dx L/4 L/4 L  L  3π 2 1 3π / 2 1  1  P  sin 2 z dz   z  2 sin 2 z  π π/2 2π  π 2 1  2 L 3L c) This is consistent with Fig. 40.5(b) since more of ψ1 is between x  and than ψ 4 4 and the proportions appear correct. 40.36: Using the normalized wave function ψ1  2 L sin(πx L) , the probabilities | ψ |2 dx are a ) (2 L) sin 2( π 4 )dx  dx L, b) (2 L) sin 2 (π 2) dx  2 dx L and c) (2 L) sin 2 (3π 4)  dx L . 2  2πx  40.37: ψ 2 ( x)   sin   L  L  2  2πx  | ψ 2 |2 dx  sin 2  dx L  L  L 2 π 2dx a) x  :| ψ 2 |2 dx  sin 2   dx  4 L  2 L L 2 b) x  :| ψ 2 |2 dx  sin 2 π  dx  0 2 L 3L 2  3π  2dx c) x  :| ψ 2 |2 dx  sin 2   dx  4 L  2 L (n  1) 2  n 2 2n  1 2 1 40.38: a) Rn     2. n2 n2 n n This is never larger than it is for n  1, and R  3. b) R approaches zero; in the classical limit, there is no quantization, and the spacing of successive levels is vanishingly small compared to the energy levels.
  11. h2 40.39: a) The transition energy E  E2  E1  (4  1) 8mL2 hc 8mcL2 8(9.11  10 31 kg)(3.00  108 m s)(4.18  10 9 m) 2 λ   E 3h 3(6.63  10 34 J  s)  λ  1.92  10 5 m. 8(9.11  10 31 kg)(3.00  108 m s)(4.18  10 9 m) 2 b) λ  34  1.15  10 5 m. (3  2 )(6.63  10 J  s) 2 2 dψ dψ π 2 dψ 40.40: a) ψ  0, so  0. b)  cos(πx L) , and as x  L,   2π 2 L3 . dx dx L L dx Clearly not. In Eq. (40.1), any point where U (x) is singular necessitates a discontinuity in dψ . dx 2  πx  40.41: a) ψ1  sin   L  L 2π 2  1  πx   2 dψ 2 π  πx  1        1   cos   dx L L L L3  2  L     dψ1 2π 2   for x  0 dx L3 2 2πx dψ 2 2 2π 2πx b) ψ 2  sin     cos L L dx L L L dψ 2 8π 2   for x  0. dx L3  πx  c) For x  L, cos    cos π  1  L dψ1 2π 2   for x  L dx L3 dψ 2 8π 2  2πx  8π 2 d)  cos   for x  L, since cos 2π  1 dx L3  L  L3 e) The slope of the wave function is greatest for ψ 2 (n  2) close to the walls of the box, as shown in Fig. 40.5.
  12.    40.42: p  pfinal  pinitial.  nπ hn p  k   . L 2L  hn ˆ At x  0 the initial momentum at the wall is pinitial   i and the final momentum, 2L  hn ˆ  hn ˆ  hn ˆ  hn ˆ after turning around, is pfinal   i . So p   i   i    i. 2L 2L  2L  L  hn ˆ At x  L the initial momentum is pinitial   i and the final momentum, after 2L  hn ˆ  hn ˆ hn ˆ hn ˆ turning around, is pfinal   i . So p   i i   i. 2L 2L 2L L
  13. d 2ψ ( x ) 40.43: a) For a free particle, U ( x)  0 so Schr  odinger' s equation becomes  dx 2 2m  Eψ ( x). h2 See graph below. b) For x < 0 ψ ( x)  e  kx . dψ ( x)  ke  kx . dx d 2ψ ( x )  k 2 e  kx . dx 2m 2 k 2 So k2   2 E  E   .  2m c) For x > 0 ψ ( x)  e  kx . dψ ( x)   ke kx . dx 2 d ψ ( x)  k 2 e kx dx 2m  2 k 2 So again k 2   2 E  E  .  2m  2 k 2 Parts (c) and (d) show ψ (x) satisfies Schr  odinger's equation, provided E  . 2m dψ ( x) d) Note is discontinuous at x  0. (That is, negative for x  0 and positive for x  dx
  14. 2 d 2ψ ( x) 40.44: b)   U ( x)ψ ( x)  Eψ ( x) 2m dx 2 d 2ψ ( x ) 2m    2 ( E  U ( x))ψ ( x) dx 2  2 d ψ ( x) 2m  2   Aψ ( x) where A   2 ( E  U ( x)). dx h Note that if E  U (x) we are inside the well and A  0. From the above equation, this d 2ψ ( x ) implies has the opposite sign of ψ (x). dx 2 If E  U (x), we are outside the well and A  0. From the above equation, this implies d 2ψ ( x ) has the same sign as ψ (x). dx 2 40.45: a) We set the solutions for inside and outside the well equal to each other at the well boundaries, x  0 and L. x  0 : A sin(0)  B  C  B  C , since we must have D  0 for x  0. 2mE L 2mE L x  L : A sin  B cos   De L since C  0 for x  L   L A sin kL  B cos kL  De  . 2mE where k   b) Requiring continuous derivatives at the boundaries yields dψ x  0:  kA cos(k  0)  kB sin(k  0)  kA  Ce k 0  kA  C dx x  L : kA cos kL  kB sin kL  De L . E  E  2m(U 0  E ) 1 40.46: T  Ge 2L with G  16 1   U  and    L U0  0   2 T  ln  . If E  5.5 eV, U 0  10.0 eV, m  9.11  10 31 kg, and T  0.0010. G 2(9.11  10 31 kg)(4.5 eV)(1.60  10 19 J eV) Then κ  34  1.09  1010 m 1 (1.054  10 J  s) 5.5 eV 5.5 eV  and G  16 1   10.0 eV   3.96  10.0 eV  1  0.0010  10 so L   1 ln    3.8  10 m  0.38 nm. 2(1.09  10 m )  3.96  10
  15. 1  (U sin hκ L) 2  40.47: a) T  1  0   4 E (U 0  E )  1 1 1  U 02 e 2 κL   U 0 e 2 κL  2 If κL  1, then sinh κL  e κL  T  1      2  16 E (U 0  E )  16 E (U 0  E )  E  E  T  Ge  2 κL where G  16 1   U .  U0  0  b) κL  1 implies either κ is large or L is large (or both are large). If L is large, the barrier is wide. If  is large, U 0  E is big, which implies E is small compared to U 0 . c) As E  U 0 , κ  0  sinh κL  κL 1 1  U 02 κ 2 L2  1  2U 0 L2 m  2  T  1      4 E (U 0  E )   4 E 2  2m(U 0  E ) since κ 2  . 2 1 1 2 U0   2mE  L  2    kL  2  But U 0  E   E  T  1   2     1     since E     2       2     2mE k2  2 .  d 2ψ 40.48: For a wave function ψ with wave number k ,  k 2ψ , and using this in dx 2 Eq. (40.1) with U  U 0  E gives k 2  2m( E  U 0 ) 2 , or k  i 2m(U 0  E )   i . 2π 2π 40.49: The angular frequency ω    12.6 rad s. Ground state harmonic T 0.500 s oscillator energy is given by 1 1 E0  ω  (1.054  10 34 J  s)(12.6 rad s)  6.62  10 34 J 2 2 6.62  10 34 J  19  4.14  10 15 eV 1.60  10 J eV  3  1  33 E  En1  En  ω  n     n     ω  E  2 E0  1.32  10 J    2  2  8.28  10 15 eV. These values are too small to be detected.
  16. 1 2 40.50: a) E  mv  (n  (1 2))ω  (n  (1 2))hf , and solving for n, 2 1 2 mv 1 (1 2)(0.020 kg)(0.360 m s) 2 1 n 2    34   1.3  1030. hf 2 (6.63  10 J  s)(1.50 Hz) 2 b) The difference between energies is ω  hf  (6.63  10 34 J  s)(1.50 Hz)  9.95  10 34 J. This energy is too small to be detected with current technology  2 d 2ψ 1 2 40.51: a) Eq. (40.21) :  k x ψ  Eψ 2m dx 2 2 d 2 Now, 2 (Cxe mω x 2  ) dx d  mω x 2 2  Cmω2 x 2 e  mω x 2   2  Ce   dx   2   2   2 xmω 4mx 4m 2 ω 2 x 3   (Ce mω x 2 )     2 2 42   mω x 2 2    3mω  mω  2 2   (Cxe )   x            3ω mω 2 2 1 2  k 3ω  Eq. (42 - 25) : ψ   x  k x   ψE but ω 2   E  .  2 2 2  m 2 b) Similar to the second graph in Fig. 40.18.
  17. 40.52: With u ( x)  x 2 , a  mω , | ψ |2 is of the form C *Cu(x)e -au(x) , which has a maximum mω 2 2 at au  1, or x  1, x    m . b) ψ and hence ψ vanish at x  0, and as x     2 ψ  0. the result of part (a) is  A 3 , found from n  1 in Eq. (40.26), and this is consistent 2 with Fig. (40.19). The figure also shows a minimum at x  0 and a rapidly decreasing ψ
  18.  2   2ψ  2 ψ  2ψ  40.53: a) Eq. (42-32):  2  2  2   Uψ  Eψ.ψ nx , ψ ny and ψ nz are all  x 2m  y z   2   d ψ nx 1 2 solutions to the 1 - D Schrodinger equation, so  k x ψ nx  Enx ψ nx , and 2m dx 2 2  2ψ  d ψ n x  2 ψ n ψ n ,  ψ 2 similarly for ψ ny and ψ nz . Now if ψ  ψ nx ( x)ψ ny ( y )ψ nz ( z ) then 2   x  dx  y z y 2    d 2ψ n y   2   ψ ψ and  ψ   d ψ nx ψ ψ . 2   dy 2  dz 2  dz 2  x y nx nz n n     Therefore:  2   2ψ  2ψ  2ψ  1  2  2  2   k ( x 2  y 2  z 2 )ψ 2m  x  y z  2    2  d 2 ψ nx  1 2      k x ψ n ψ n ψ n  2m  dx 2  2 x  y z       2  d 2ψ n y  1 2      k y ψ ψ ψ  2m  dy 2  2 ny  nx nz       2  d 2ψ n  1 2    z   k z ψ nz ψ nx ψ n y   2m  dz 2  2      [ Enx  Eny  Enz ]ψ nx ψ ny ψ nz  [ Enx  Eny  Enz ]ψ  Enx n y nz ψ  1  1  1   Enx ny nz   nx     n y     nz   ω  2  2  2   3  Enx ny nz  nx  n y  n z   ω.  2 b) Ground state energy E 000  3 ω. 2 5 First excited state energy E100  E001  E010  ω. 2 c) As seen in b) there is just one set of quantum numbers (0, 0, 0) for the ground state and three possibilities (1, 0, 0), (0, 1, 0) and (0, 0, 1) for the first excited state.
  19. 40.54: Let 1  k1 m , ω2  k 2 m , ψ nx ( x) be a solution of Eq. (40.21) with Enx    1  nx   ω1 , ψ nx ( y ) be a similar solution, and ψ nz ( z ) be a solution of Eq.(42 - 25) but  2  1 with z as the independent variable instead of x, and energy Enz   nz  ω2. (a) As in  2 Problem 40.53, look for a solution of the form ψ ( x, y, z )  ψ nx ( x)ψ ny ( y )ψ nz ( z ). Then, 2  2ψ  1     Enx  k1x 2 ψ 2m x 2  2  ψ 2 ψ2 with similar relations for and 2 . Adding, y 2 z 2   2ψ  2ψ  2ψ     1 1 1   2  2  2    Enx  En y  Enz  k1x 2  k1 y 2  k 2 z 2 ψ  x  2m  y z   2 2 2   ( Enx  Eny  Enz  U )ψ  ( E  U )ψ where the energy E is   1 2 E  Enx  En y  Enz  (n x  n y  1)ω12   n z  ω2 ,   2  nx , n y and nz all nonnegative integers. b) The ground level corresponds to nx  n y  nz  0, and E  (ω 2  ω 2 2). The first excited level corresponds to nx  n y  0 and 1 2 nz  1, as ω 2  ω 2 , and E  (ω 2 (3 2)ω 2 ). There is only one set of quantum numbers 2 1 1 2 for both the ground state and the first excited state.
  20. 40.55: a) ψ ( x)  A sin kx and ψ ( L 2)  0  ψ ( L 2)   kL   kL 2nπ 2π  0  A sin    nπ  k    2  2 L λ L h nh p2 n 2 h 2 ( 2n ) 2 h 2 λ  pn    En  n   , where n λn L 2m 2mL2 8mL2 n  1, 2... b) ψ ( x)  A cos kx and ψ ( L 2)  0  ψ (  L 2)  kL  kL π (2n  1)π 2π  0  A cos     (2n  1)  k    2 2 2 L λ 2L (2n  1)h λ  pn  (2n  1) 2L (2n  1) 2 h 2  En  n  0, 1, 2... 8mL2 c) The combination of all the energies in parts (a) and (b) is the same energy n2h2 levels as given in Eq. (40.9), where En  . 8mL2 d) Part (a)’s wave functions are odd, and part (b)’s are even. 40.56: a) As with the particle in a box, ψ ( x)  A sin kx, where A is a constant and k 2  2mE 2 . Unlike the particle in a box, however, k and hence E do not have simple forms. b) For x  L, the wave function must have the form of Eq. (40.18). For the wave function to remain finite as x  , C  0. The constant κ 2  2m(U 0  E ) , as in Eq. (14.17) and Eq. (40.18). c) At x  L, A sin kL  De  κL and kA cos kL   κDe  κL . Dividing the second of these by the first gives k cot kL  κ , a transcendental equation that must be solved numerically for different values of the length L and the ratio E U 0 .
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