Physics exercises_solution: Chapter 39

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Physics exercises_solution: Chapter 39

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Bộ tài liệu bài tập tham khảo môn vật lý bậc đại học bằng tiếng anh Chapter 39

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Nội dung Text: Physics exercises_solution: Chapter 39

  1. h h (6.63  10 34 J  s) 39.1: a) λ   p   10  2.37  10 24 kg  m s . p λ (2.80  10 m) p 2 (2.37  10 24 kg  m s) 2 b) K    31  3.08  10 18 J  19.3 eV. 2m 2(9.11  10 kg) h h 39.2: λ   p 2mE (6.63  10 34 J  s)   7.02  10 15 m.  27 19 2(6.64  10 kg ) (4.20  10 eV) (1.60  10 J e V) 6 h 6.63  1034 J  s 39. 3: a) λ e    31  1.55  1010 m. me v (9.11  10 kg) (4.70  10 m s) 6 me  9.11  10 31 kg  b) λ p  λe    1.67  10  27 kg  1.55  10 m  8.46  10 m.  10 14 mp   15 hc (4.136  10 eV  s) (3.00  108 m s) 39.4: a) E    6.2 keV. λ (0.20  10 9 m) p 2 (h λ ) 2 ((6.626  10 34 J  s) (0.20  10 9 m)) 2 b) K    2m 2m 2(9.11  10 31 kg)  6.0  10 18 J  37 eV. Note that the kinetic energy found this way is much smaller than the rest energy, so the nonrelativistic approximation is appropriate. p 2 (h λ) 2 ((6.626  10 34 J  s) (0.20  10 9 m)) 2 c) K     27  8.3  10 22 J  2m 2m 2(6.64  10 kg) 5.2 meV. Again, the nonrelativistic approximation is appropriate. nh 39.5: a) In the Bohr model mv rn  . The de Broglie wavelength is 2π h h 2π rn λ   for n  1 : r1  a0  5.29  10 11 m  λ1  2π (5.29  10 11 m)  3.32  10 10 m. p mv n This equals the orbit circumference. 2π (16a0 ) b) n  4 : r4  (4) 2 a0  16a0  λ 4   4 λ, 4  λ 4  1.33  109 m. The de Broglie wavelength is a quarter of the circumference of the orbit, 2πr4 . p2 39.6: a) For a nonrelativistic particle, K  , so 2m h h λ  . p 2 Km b) (6.63  10 34 J  s) 2(800 eV) (1.60  1019 J/eV ) (9.11  10 31 Kg )  4.34  10 11 m.
  2. h h 6.63  1034 J  s 39.7: λ    3.90  10 34 m. p mv (0.005 kg) (340 m s) We should not expect the bullet to exhibit wavelike properties. 39.8: Combining Equations 37.38 and 37.39 gives p  mc γ 2  1. h a) λ   (h mc)  2  1  4.43  10 12 m. (The incorrect nonrelativistic calculation p gives 5.05  10 12 m.) b) (h mc)  2  1  7.07  10 13 m. 39.9: a) photon hc hc (6.626  10 34 J  s) (2.998  108 m s) E so λ    62.0 nm λ E (20.0 eV) (1.602  10 19 J e V) electron E  p 2 (2m) so p  2mE  2(9.109  10 31 kg ) (20.0 eV) (1.602  10 19 J e V)  2.416  10 24 kg  m s λ  h p  0.274 nm b) photon E  hc λ  7.946  10 19 J  4.96 eV electron λ  h p so p  h λ  2.650  10 27 kg  m s E  p 2 (2m)  3.856  10 24 J  2.41  10 5 eV c) You should use a probe of wavelength approximately 250 nm. An electron with λ  250 nm has much less energy than a photon with λ  250 nm, so is less likely to damage the molecule. h h 39.10: λ v mv mλ 2 1 1  h  h2 K  mv 2  m    2 2  mλ  2mλ 2 They will not have the same kinetic energy since they have different masses.  h2    K p  2mp λ  me 2   K e  h  mp2   2m λ 2   e  9.11  10 31 kg   27  5.46  10 4 1.67  10 kg
  3. 39.11: a) λ  0.10 nm p  mv  h λ so v  h ( mλ )  7.3  10 6 m s b) E  1 mv 2  150 eV 2 c) E  hc λ  12 keV d) The electron is a better probe because for the same λ it has less energy and is less damaging to the structure being probed. 39.12: (a) λ  h mv  v  h m λ Energy conservation: eV  1 mv 2 2 mv 2 m( mλ ) 2 h h2 V    2e 2e 2emλ 2 (6.626  10 34 J  s) 2  2(1.60  10 19 C) (9.11  10 31 kg ) (0.15  10 9 m) 2  66.9 V hc (6.626  10 34 J  s) (3.0  108 m s) (b) Ephoton  hf    1.33  10 15 J λ 0.15  10  9 m eV  K  E photon E photon 1.33  10 15 J V    8310 V e 1.6  10 19 C 39.13: For m =1, h λ  d sin θ  2mE h2 (6.63  10 34 J  s) 2 E  2md 2 sin 2 θ 2(1.675  10 27 kg ) (9.10  10 11 m) 2 sin 2 (28.6)  E  6.91  10 20 J  0.432 eV. 39.14: Intensity maxima occur when d sin θ  mλ. h h mh λ  so d sin θ  . p 2ME 2 ME (Careful! Here, m is the order of the maxima, whereas M is the mass of the incoming particle.) mh (2) (6.63  10 34 J  s) a) d   2ME sin θ 2(9.11  10 31 kg) (188 eV) (1.60  10 19 J e V) sin(60.6)  2.06  10 10 m  0.206 nm. b) m = 1 also gives a maximum.  (1) (6.63  10 34 J  s)  θ  arcsin    2(9.11  10 31 kg) (188 eV) (1.60  10 19 J e V) (2.06  10 10 m)     25.8. This is the only other one. If we let m  3, then there are no more maxima.
  4. m2h2 (1) 2 (6.63  10 34 J  s) 2 c) E   2Md 2 sin 2 θ 2(9.11  10 31 kg) (2.60  10 10 m) 2 sin 2 (60.6)  7.49  10 18 J  46.8 eV. Using this energy, if we let m  2, then sin θ  1. Thus, there is no m  2 maximum in this case. 39.15: Surface scattering implies d sin θ  mλ. If m  1 : θ  arcsin[λ d ] h h 6.63  10 34 J  s But λ    p 2mE 2(6.64  10 27 kg ) (840 eV) (1.60  10 19 J/eV)  λ  4.96  10 13 m.  4.96  10 13 m  So θ  arcsin  11   0.341.  8.34  10 m  h h  mh  39.16: The condition for a maximum is d sin θ  mλ.  λ, so θ  arcsin  . p Mv  dMv  (Careful! Here, m is the order of the maximum, whereas M is the incoming particle mass.)  h  a) m  1  θ1  arcsin    dMv   6.63  10 34 J  s   arcsin  (1.60  10 m) (9.11  10 kg) (1.26  10 m s)   2.07. 6 31 4     (2) (6.63  10 34 J  s)  m  2  θ2  arcsin  (1.60  10 6 m) (9.11  10 31 kg ) (1.26  10 4 m s)      4.14. b) For small angles (in radians!) y  Dθ , so  π radians  y1  (50.0 cm) (2.07)    1.81 cm.  180   π radians  y2  (50.0 cm) (4.14)    3.61 cm  180   y2  y1  3.61 cm  1.81 cm  1.81 cm. h h h (6.63  10 34 J  s) 39.17: a) px   mv x x   v x   2π 2π 2πmx 2π (1200 kg ) (1.00  10 6 m)  8.79  10 32 m s . b) Knowing the position of a macroscopic object (like a car) to within 1.00 μm is, for all practical purposes, indistinguishable from knowing “exactly” where the object is. Even with this tiny position uncertainty of 1.00 μm , the velocity uncertainty is insanely small by our standards.
  5. h h h 39.18: a) p y y  for minimum uncertainty  mv y y   v y  . 2π 2π 2πmy (6.63  10 34 J  s)   3.2  10 4 m s 2π (1.67  10 27 kg) (2.0  10 12 m) b) For minimum uncertainty, h h (6.63  10 34 J  s) z    31  4.63  10 4 m. 2πp z 2πmv z 2π (9.11  10 kg) (0.250 m s) 39.19: Heisenberg’s Uncertainty Principles tells us that: h xp x  . We can treat the standard deviation as a direct measure of uncertainty. 2π h Here xp x  (1.2  10 10 m) (3.0  10 25 kg  m s)  3.6  10 35 J  s but  1.05  10 34 J  s 2π h Therefore xp x  so the claim is not valid . 2π 39.20: a) (x) (mv x )  h 2π , and setting vx  (0.010)vx and the product of the uncertainties equal to h / 2π (for the minimum uncertainty) gives v x  h (2πm(0.010)x)  57.9 m s . b) Repeating with the proton mass gives 31.6 mm s. h h (6.63  10 34 J  s) 39.21: px   p    9.82  10 25 kg  m s. 2π 2πx  0.215  10 9 m  2π      2  25 p 2 (9.82  10 m) 2 b) K    26  4.95  10 24 J  3.09  10 5 eV. 2m 2(9.75  10 kg ) 1.00 kg c) K total  NK  (4.95  10 24 J)  50.8 J. 9.75  10 26 kg Ni K total 50.8 J d) mgh  K total  h    5.18 m. mg (1.00 kg) (9.81 m s 2 ) e) One is claiming to know both an exact momentum for each atom (giving rise to an exact kinetic energy of the system) and an exact position of each atom (giving rise to an exact potential energy of the system), in violation of Heisenberg’s uncertainty principle. h 39.22: Et  E  mc 2 m  2.06  10 9 eV c 2  3.30  10 10 J c 2 . 2π h 6.63  10 34 J  s t   10  3.20  10 25 s. 2πmc 2 2π (3.30  10 J)
  6. h h (6.63  10 34 J  s) 39.23: Et   E    21  1.39  10 14 J  8.69  2π 2πt 2π (7.6  10 s) 10 4 eV  0.0869 MeV. E 0.0869 MeV c 2  2  2.81  10 5. E 3097 MeV c h (6.63  10 34 J  s) 39.24: E   3  2.03  10 32 J  1.27  10 13 eV. 2πt 2π (5.2  10 s) 39.25: To find a particle’s lifetime we need to know the uncertainty in its energy. E  (m)c 2  (0.145) (4.5) (1.67  10 27 kg) (3.00  108 m s) 2  9.81  10 11 J h 6.63  10 34 J  s  t    1.08  10 24 s. 2πE 2π (9.81  10 J) 11 p 2 (h λ) 2 (h λ) 2 39.26: a) eV  K   , so V   419 V. 2m 2m 2me b) The voltage is reduced by the ratio of the particle masses, 9.11  10 31 kg (419 V)  0.229 V. 1.67  10 27 kg h h 39.27: a) We recall λ   . But the energy of an electron accelerated through a p 2mE potential is just E  eV h (6.63  10 34 J  s)  λe   2mV 2(9.11  10 31 kg ) (1.60  10 19 C) (800 V)  λ e  4.34  10 11 m. b) For a proton, all that changes is the mass, so 9.11  10 31 kg λp  me mp λe   27 1.67  10 kg  . 4.34  10 11 m   λ e  1.01  10 12 m. 39.28:    ψ  sin ωt , so 2 2    *  ψ *ψ sin 2 ωt  ψ sin 2 ωt . 2  is not time-independent, so  is not the wavefunction for a stationary state. 39.29: a) ψ  x   A sin kx . The probability density is ψ  A2 sin 2 kx, and this is greatest when 2 nπ sin 2 kx  1  kx  , n  1, 3, 5 . . . 2 nπ nπ nλ x   , n  1, 3, 5 . . . 2k 2(2π λ) 4
  7. 2 b) The probability is zero when ψ  0 , which requires nπ nλ sin 2 kx  0  kx  nπ  x   , n  0, 1, 2 ... k 2 39.30: a) The uncertainty in the particle position is proportional to the width of ψ x  , and is inversely proportional to  . This can be seen by either plotting the function for different values of  , finding the expectation value x 2   ψ 2 x 2 dx for the normalized wave function or by finding the full width at half-maximum. The particle’s uncertainty in position decreases with increasing  . The dependence of the expectation value  x 2  on  may be found by considering  x e 2  2x 2 dx x   2   e  2x 2 dx  1    2x 2   2   = ln  e dx    1   1   1  2  ln  e u du   , 2   2   4 where the substitution u  x has been made. (b) Since the uncertainty in position decreases, the uncertainty in momentum must increase.  x  iy   x  iy  39.31: f ( x, y )    x  iy  and f ( x, y )   x  iy   *       2  x  iy   x  iy   f  f f*  x  iy    x  iy   1.        39.32: The same. 2 ψ ( x, y, z )  ψ * ( x, y, z )ψ ( x, y, z ) 2 ψ ( x, y , z )e i  (ψ * ( x, y, z )e  i )(ψ ( x, y, z )e  i )  ψ * ( x, y, z )ψ ( x, y, z ). The complex conjugate means convert all i ’s to  i ’s and vice-versa. e i  e  i  1. 39.33: Following the hint: If we Taylor expand sin( ax ) about a point x0 , we get f ( x0 )  f ( x0 )( x  x0 )      sin(ax0 )  a cos(ax0 )( x  x0 )    . If x  x0 is small we can even ignore the first order term and sin( ax )  sin ( ax0 ). 32 2  πx  For us x  x0  0.01L which is small compared to L so ψ ( x, y, z )    sin  0   L  L   πy   πz   sin  0  sin  0 .  L   L 
  8. L a) x0  y0  z0  4 6 2  2 3 3 2 π P   ψ dV    sin 6   V      2 6  2  (0.01L)  1.00  10 . 3 L 4  L   1 b) x0  y 0  z 0  2 3 3 2 π 2 P   ψ dV    sin 6   V    (0.01 L) 3  8.00  10 6. 2  L 2  L  2 d 2ψ 39.34: Eq. (39.18):  Uψ  Eψ 2m dx 2 Let ψ  Aψ1  Bψ 2  2 d 2  ( Aψ1  Bψ 2 )  U ( Aψ1  Bψ 2 )  E ( Aψ1  Bψ 2 ) 2m dx 2  2 d 2ψ1   2 d 2ψ 2   A   2m dx 2  Uψ1  Eψ1   B    2m dx 2  Uψ 2  Eψ 2   0.      But each of ψ1 and ψ 2 satisfy Schröedinger’s equation separately so the equation still holds true, for any A or B. 2 d 2ψ 39.35:   Uψ  BE1ψ1  CE2ψ 2 . 2m dx 2 If ψ were a solution with energy E, then BE1ψ1  CE2ψ 2  BEψ1  CEψ 2 or B( E1  E )ψ1  C ( E  E2 )ψ 2 . This would mean that ψ1 is a constant multiple of ψ 2 , and ψ1 and ψ2 would be wave functions with the same energy. However, E1  E 2 , so this is not possible, and ψ cannot be a solution to Eq. (39.18). h (6.63  10 34 J  s) 39.36: a) λ   2mK 2(9.11  10 31 kg)(40 eV)(1.60  10 19 J eV)  1.94  10 10 m. R R (2.5m)(9.11  10 31 kg)1 2 b)    6.67  10 7 s. v 2E m 19 2(40eV)(1.6  10 J eV) λ c) The width w is w  2 R ' and w  v y t  p y t m, where t is the time found in part (b) a 2mλR and a is the slit width. Combining the expression for w, p y   2.65  10  28 kg  m s. at h d) y   0.40 μm, which is the same order of magnitude. 2πp y
  9. 39.37: a) E  hc λ  12eV b) Find E for an electron with λ  0.10  10 6 m. λ  h p so p  h λ  6.626  10 27 kg  m s E  p 2 (2m)  1.5  10 4 eV E  qV so V  1.5  10 4 V v  p m  (6.626  10 27 kg  m s) (9.109  10 31 kg)  7.3  103 m s c) Same λ so same p. E  p 2 (2m) but now m  1.673  1027 kg so E  8.2  108 eVand V  8.2  108 V. v  p m  (6.626  1027 kg  m s) (1.673  1027 kg)  4.0 m s 39.38: (a) Single slit diffraction: a sin θ  mλ λ  a sin θ  (150  10 9 m)sin20  5.13  10 8 m λ  h mv  v  h mλ 6.626  1034 J  s v  31 8  1.42  104 m s (9.11  10 kg)(5.13  10 m) (b) a sin θ2  2λ λ  5.13  10 8 m  sin θ2   2   2  150  10 9 m    0.684  a   θ2  43.2 39.39: a) The first dark band is located by sin θ  λ a λ 150nm a   355nm sin θ sin25.0 b) Find λ for the electrons hc E  1.324  10 18 J λ photon E  p 2 (2 m) so p  2mE  1.553  10 24 kg  m s λ  h p  4.266  10 10 m No electrons at locations of minima in the diffraction pattern. The angular position of these minima are given by: sin θ  mλ a  m(4.266  10 10 m) (355  10 9 m)  m(0.00120), m  1,  2,  3, ... m  1,   0.0689; m  2,   0.138; m  3,   0.207; ... 39.40: According to Eq. 35.4 d sin θ (40.0  10 6 m)sin(0.0300) λ   600 nm. m 2 The velocity of an electron with this wavelength is given by Eq. 39.1 p h (6.63  10 34 J  s) v   31 9  1.21  103 m s . m mλ (9.11  10 kg)(600  10 m) Since this velocity is much smaller than c we can calculate the energy of the electron classically 1 1 K  mv 2  (9.11  10 31 kg)(1.21  10 3 m s) 2  6.70  10  25 J  4.19 μeV. 2 2
  10. 39.41: The de Broglie wavelength of the blood cell is h (6.63  10 34 J  s) λ  14 3  1.66  10 17 m. mv (1.00  10 kg)(4.00  10 m s) We need not be concerned about wave behavior. 12  v2  h1  2   c  h 39.42: a) λ    p mv  v2  h 2v 2  λ 2 m 2 v 2  h 2 1  2   h 2  2  c    c v2  λ 2 m2v 2  h 2 2  h 2 c 2 h c2  v2   2 2 2  2 2 h2   λ m c  λ m  2      h 2  1   c    c v 12 .   mcλ  2  1      h      c  1  mcλ  2  b) v   c1     (1  )c.   λ 2  12  2 h    1        (h mc)       m2c 2λ 2  . 2h 2 h c) λ  1.00  10 15 m  . mc (9.11  10 31 kg) 2 (3.00  108 m s) 2 (1.00  10 15 m) 2 So   34  8.50  10 8 2(6.63  10 J  s) 2  v  (1  Δ)c  (1  8.50  10 8 )c. h h h 39.43: a) Recall λ    . So for an electron: p 2mE 2mqV 6.63  10 34 J  s λ  λ  1.10  10 10 m. 31 19 2(9.11  10 kg)(1.60  10 C)(125 V) b) For an alpha particle: 6.63  10 34 J  s λ  9.10  10 13 m.  27 19 2(6.64  10 kg)2(1.60  10 C)(125 V)
  11. 39.44: a) E 2  p 2 c 2  m 2 c 4 and E  K  mc 2  ( K  mc 2 ) 2  p 2 c 2  m 2 c 4 [( K  mc 2 ) 2  m 2 c 4 ]1 2 [ K 2  2 Kmc 2  m 2 c 4  m 2 c 4 ]1 2  p  c c [ K ( K  2mc )]2 12  c h hc λ  . p [ K ( K  2mc 2 )]1 2 hc h b) i) K  mc 2 λ  2 12  . (2 Kmc ) (2 Km)1 2 hc hc ii) K  mc 2 λ  2 12  . (K ) K 9 c) K  7.00  10 eV  1.12  10 J. 9 m  1.67  10 27 kg. hc λ [ K(K  2mc 2 )]1 2 (6.63  10 34 J  s)(3.00  10 8 m s)  [(1.12  10 9 J)(1.12  10 9 J  2(1.67  10  27 kg)(3.00  10 8 m s) 2 )]1 2  1.57  10 16 m. d) K  25.0  10 6 eV  4.00  10 12 J. m  9.11  10 31 kg. (6.63  10 34 J)(3.00  108 m s) λ [4.00  10 12 J(4.00  10 12 J  2(9.11  10 31 kg)(3.00  108 m s)]1 2  4.87  10 14 m. 39.45: a) Since K  mc 2 we must use the relativistic expression for energy. E 2  p 2c 2  m 2c 4 but E  K  mc 2  ( K  mc 2 ) 2  p 2c 2  m 2c 4 [( K  mc 2 ) 2  m 2 c 4 )]1 2 h hc  p λ  . c p [( K  mc )  m 2 c 4 ]1 2 2 2 hc h If K  3mc 2 then λ   . [(4mc )  m c ] 2 2 2 4 12 15mc b) i) m  9.11  10 31 kg K  3mc 2  3(9.11  10 31 kg)(3.00  10 8 m s) 2  2.46  10 13 J  1.54 MeV. h (6.63  10 13 J  s) λ  15mc 15 (9.11  10 31 kg)(3.00  108 m s)  6.2  10 13 m.
  12. ii) m  1.67  10 27 kg K  3mc 2  3(1.67  10 27 kg)(3.00  10 8 m s) 2  4.51  10 10 J  2.82  103 MeV. h (6.63  10 34 J  s) λ  15mc 15 (1.67  10 27 kg)(3.00  108 m s)  3.42  10 16 m. h (6.626  10 34 J  s) 39.46: p ~  10  2.0  10 24 kg  m s , which is comparable to 2πa0 2π (0.5292  10 m) mv1  2.0  1024 kg  m s. h h h p 39.47: x  0.40λ  0.40 . But xp x   p x (min)    0.40 p. p 2π 2πx 2π (0.4) (6.626  10 34 J  s) 39.48: a)  2.1  10 20 kg  m s. 2π (5.0  10 15 m) b) K  ( pc) 2  (mc 2 ) 2  mc 2  1.3  10 13 J  0.82 MeV. c) The result of part (b), about 1 MeV  1  10 6 eV , is many orders of magnitude larger than the potential energy of an electron in a hydrogen atom. h 6.63  10 34 J  s 39.49: a) p (min)    2.1  10 20 kg  m s 2πx 2π (5.0  10 15 m) b) E  ( pc) 2  (mc 2 ) 2  [(2.1  10 20 kg  m s)(3.0  10 8 m s)]2  [(9.11  10 31 kg)(3.0  10 8 m s) 2 ]2  6.3  10 12 J  39.5MeV. K  E  mc 2  38.8 MeV q1q2 (1.60  10 19 C) 2 c) The coulomb potential energy is U  U    4πε0V 4πε0 (5.0  10 15 m)  4.60  10 14 J  0.29 MeV Hence there is not enough energy to “hold” the electron in the nucleus. 39.50: a) Take the direction of the electron beam to be the x-direction and the direction of motion perpendicular to the beam to be the y-direction. Then, the uncertainty r in the position of the point where the electrons strike the screen is p y x r  v y t  m vx h x  2πmy 2K m  9.56  10 10 m, which is (b) far too small to affect the clarity of the picture.
  13. h (6.63  10 34 J  s) 39.51: E   17  1.26  10 18 J 2πt 2π (8.4  10 s) E 1.26  10 18 J But E  (m)c  m  2  2  1.4  10 35 kg c (3.0  10 m s) 8 2 m 1.4  10 35 kg    31  5.8  10 8 m 264(9.11  10 kg) h h h h 39.52: a) λ     . p 2mE 3 3mkT 2m( kT ) 2 b) We would roughly expect the length scale of the problem to go like V 1 3 (e.g., for a cube 4 3 V  l 3 so l  v1 3 and for a sphere V  πR 3 , so R  ( V 1 3  V 1 3 ). Let n be the number of 3 4π molecules along one length (again, think of a cube) so that n 3  N , the total number of particles 13 l V  in the volume. So n  N 1 3 . Thus, the typical spacing between particles is    . The exact n N relationship will change depending on the geometry, but the scaling is correct up to a multiplicative constant. 13 13 V  h V  Nh 3 c) λ       V  . N 3mkT  N  3mkT 3 2  6.02  10 particles mol 1.00 mol 6.63  10 J  s  23 34 3  3.03  10 8 m 3 d) Vwave 32.66  10 kg 1.38  10 J K  293 K   26  23 32  30.3 mm3 . NkT PV  NkT  VSTP  P     6.02  10 particles mol 1.00 mol 1.38  10  23 J K 293 K  23   1.01  105Pa   VSTP  0.0241 m 3  2.41  10 7 mm 3 .  VSTP is far larger than Vwave so the wave nature is not important. 1.00 kg e) N   5.59  10 24 Ag atoms( conduction e). 1.79  10 25 kg atom (5.59  1024 electrons)(6.63  1034 J  s)3 Vwave   1.40 m 3 . [(3)(9.11 10 31 kg)(1.38  10 23 J K)(293 K)]3 2 1.00 kg VReal   9.52  10 5 m 3 . 1.05  10 kg m 4 3 The real volume is much smaller than the wave limit volume. So, the wave nature of the electrons must be accounted for.
  14. h h 6.63  10 34 J  s 39.53: a) λ  v   1.1  10 35 m s. mv mλ (60 kg)(1.0 m) d 0.80 m b) t    35  7.2  1034 s  2.3  10 27 years. v 1.1  10 m s Therefore, we will not notice diffraction effects while passing through doorways. hc 39.54: a) E  2.58 eV  4.13  10 19 J, with a wavelength of λ   4.82  10 7 m  E 482 nm. h (6.63  10 34 J  s) b) E   7  6.43  10 28 J  4.02  10 9 eV. 2πt 2π (1.64  10 s) c) λE  hc, so (λ ) E  λE  0, and E E  λ λ , so λ  λ E E   6.43  10 28 J  (4.82  10 m)  7  4.13  10 19 J   7.50  10 m  7.50  10 nm.  16 7   h 22 6.63  10 J  s 39.55: E    4.71  10 20 J  0.294 eV. 2πt 2π (2.24  10 3 s) Note that this uncertainty is much larger than the real uncertainty as compared to the 4.50 eV. λ 39.56: sin θ   sin θ , and λ  (h p)  (h 2mE ) , and so λ  h  θ   arcsin sin θ   λ 2mE    (6.63  10 34 J  s) sin 35.8   arcsin    (3.00  10 11 m) 2(9.11  10 31 kg)(4.50  10 3 )(1.60  10 19 J eV)     20.9. 39.57: a) The maxima occur when 2d sin θ  mλ as described in Section 38.7. h h (6.63  10 34 J  s) b) λ     1.46  p 2mE  2(9.11  10 37 kg )(71.0 eV) 1.60  10 19 J eV   mλ  10 10 m  0.146 nm. θ  sin 1   ( Note : This m is the order of the maximum, not the mass.)  2d   (1)(1.46  1010 m)   sin 1   2(9.10  1011 m)   53.3.    c) The work function of the metal acts like an attractive potential increasing the kinetic energy of incoming electrons by e . An increase in kinetic energy is an increase in momentum that leads to a smaller wavelength. A smaller wavelength gives a smaller angle  (see part (b)).
  15. 39.58: a) Using the given approximation, E  1 2   (h x) 2 m  kx 2 , (dE dx)  kx  (h 2 mx 3 ), h and the minimum energy occurs when kx  (h 2 mx 3 ), or x 2  . The minimum energy is mk then h k m. b) They are the same. dU 39.59: a) U  A x but F   . For x  0, x  x  F   A. For x  0, x  dx Ax  x  F  A. So F ( x)   for x  0. x p2 h2 b) From Problem 39.58, E  K  U   A x , and px  h  E   A x. 2m 2mx 2 h2 dE dE For x  0; E  2  Ax. The minimum energy occurs when 0 0 2mx dx dx 13 13 13 h2  h2  h2  h2  3  h 2 A2   3  A  x    . So Emin   mA   A  mA      mx   2m(h 2 mA) 2 3   2 m    39.60: For this wave function,    ψ1e iω1t ψ 2 e iω2t , so   2     (ψ1 e iω1t  ψ 2 e iω2t )(ψ1e iω1t  ψ 2 e iω2t )    ψ1ψ1  ψ 2ψ 2  ψ1 ψ 2 e i ( ω1 ω2 )t  ψ 2ψ1e i ( ω2 ω1 ) t .  *  2 The frequencies ω1 and ω2 are given as not being the same, so  is not time-independent, and  is not the wave function for a stationary state. 39.61: The time-dependent equation, with the separated form for  ( x, t ) as given becomes  2 d 2ψ  iψ (iω)     2m dx 2  U ( x)ψ .    Since ψ is a solution of the time-independent solution with energy E , the term in parenthesis is Eψ , and so ω  E , and ω  ( E ). 2π E E 39.62: a) ω  2π f   . h  2π 2π p k  p . λ h  p 2 (k)2 ω  E  KE   2m 2m k 2 ω . 2m
  16. 2  2ψ ( x, t ) b) From Problem 39.61 the time-dependent Schrödinger’s equation is   2m x 2 ψ ( x, t ) U ( x)ψ ( x, t )  i . U ( x)  0 for a free particle, so t  2ψ ( x, t ) 2mi ψ ( x, t )  . x 2  t Try ψ ( x, t )  cos(kx  ωt ) ψ ( x, t )  Aω sin( kx  ωt ) t ψ ( x, t )  2ψ   Ak sin( kx  ωt ) and 2  Ak 2 cos(kx  ωt ). x x  2mi  Putting this into the Schrödinger’s equation, Ak 2 cos(kx  ωt )    Aω sin(kx  ωt ).    This is not generally true for all x and t so is not a solution. c) Try ψ ( x, t )  A sin( kx  ωt ) ψ ( x, t )   Aω cos(kx  ωt ) t ψ ( x, t )  Ak cos(kx  ωt ) x  2 ψ ( x, t )   Ak 2 sin( kx  ωt ). x 2  2mi  Again,  Ak 2 sin( kx  ωt )    Aω cos(kx  ωt ) is not generally true for all x and    t so is not a good solution. d) Try ψ ( x, t )  A cos(kx  ωt )  B sin( kx  ωt ) ψ ( x, t )   Aω sin(kx  ωt )  Bω cos(kx  ωt ) t ψ ( x, t )   Ak sin( kx  ωt )  Bk cos(kx  ωt ) x  2ψ ( x, t )   Ak 2 cos((kx  ωt )  Bk 2 sin( kx  ωt ). x 2 Putting this into the Schrödinger’s equation, 2mi  Ak 2 cos(kx  ωt )  Bk 2 sin( kx  ωt )   ( Aω sin( kx  ωt )  Bω cos(kx  ωt )).  k 2 Recall that ω  . Collect sin and cos terms. ( A  iB )k 2 cos(kx  ωt )  (iA  B)k 2 sin (kx  2m ωt )  0. This is only true if B = iA.
  17. 39.63: a) The ball is in a cube of volume 125 cm3 to start with, and hence has an uncertainty of 5 cm in any direction. x  0.05 m. (The x-direction in the horizontal, side-to-side direction.) h 0.0663 J. Now p x    0.21 kg  m s. 2πx 2π (0.05m) 12m b) The time of flight is t   2.0 s. 6.0 m s So the uncertainty in the x-direction at the catcher is  p   0.21 kg  m s  x  (v)t   t    (2.0 s)  x  1.7 m.   m   0.25 kg  2  βy 2  γz 2 ) 2 39.64: a) ψ 2  A2 x 2 e 2 ( αx . To save some algebra, let u  x 2 , so that ψ   2 2 1 ue  2u f ( y, z ), and ψ  (1  2u ) ψ ; the maximum occurs at u 0  , u 2 1 x0   . 2 b) ψ vanishes at x  0, so the probability of finding the particle in the x  0 plane is zero. The wave function vanishes for x  . 2 2 2 39.65: a) ψ ( x, y, z )  Ae  ( x  y  z ) but the distance to the origin is just r  x 2  y 2  z 2 . So, doing a change of variable. 2 ψ (r )  Ae r . However, the probability is the integral of this function over all space. It is best to think of the volume integral in spherical coordinates where dV  dx dy dz  r 2 sin θ dr dθ dω Now 2π  2 2r 2 p   d  sin θdθ  ψ (r ) r dr  4π A 2 2  r 2 dr. Thus we interpret the probability of 0 0 e 2 2 finding a particle in a spherical shell between r and r+dr as 4π A r 2 e 2r dr. d 2 2 2 2 2 b) (4π A r 2 e  2r )  0  4π A (2re 2r  4r 3e 2r )  0 dr 1 2 r . This is not the same as the maximum value of just ψ (r ) (which has a 2 maximum only at r = 0). The “extra” r 2 from looking at the function in spherical coordinates (that is, having the variable be “distance from the origin” rather than the cartesian coordinates) makes all the difference. 2 2 39.66: a) B(k )  e  k B (0)  Bmax  1 1 2 2 B(k h )   e  kh  ln (1 2)   2 k h 2 2 1  kh  ln (2)  ωk . 
  18.  π  x 2 / 4 2 Using tables: (b) ψ ( x)   e  2 2 k cos kxdk  (e ). 0 2 ψ (x) is a maximum when x = 0. π  xh / 4 2 2 1  xh 2 c) ψ ( xh )  when e    ln (1/ 2) 4 2 4 2  xh  2 ln 2  ωx  hω  d) ω p ωx   k ωx  h 1  2π    ln 2  2 ln 2  h (2ln 2)   h ln 2 .  2π   2π π k0  k0 1 sin kx sin k 0 x 39.67: a) ψ ( x)   B(k ) cos kxdk   ( ) cos kxdk   0 0 k0 k0 x 0 k0 x π b) ψ (x) has a maximum value at the origin x  0. ψ ( x0 )  0 when k 0 x0  π so x0  . Thus k0 2π the width of this function wx  2 x0  . k0 2π If k 0  , wx  L. L  c) If k 0  w x  2 L. L
  19.  hw   2π  hwk hk 0 d) w p wx   k       h.  2π   k0  k0 k0 h The uncertainty principle states that w p wx  . For us, no matter what 2π h k 0 is, w p w x  h, which is greater than . 2π 39.68: a) For a standing wave, nλ  2 L, and p 2 (h λ ) 2 n 2 h 2 En   . 2m 2m 8mL2 b) With L  a0  0.5292  10 10 m, E1  2.15  10 17 J  134 eV. 2y 39.69: Time of flight of the marble, from free-fall kinematic equation is just t   g 2(25.0 m)  2.26 s 9.81 m s 2   px  ht x f  xi  (v x )t  xi   t   xi  m  2xi m to minimize x f with respect to xi d (x f )  ht  ht  0  1  xi (min)    d (xi ) 2πm(xi ) 2  2πm  ht ht 2ht  x f (min)    2πm 2πm πm 2(6.63  10 34 J  s)(2.26 s) x f (min)   2.18  10 16 m  2.18  10 7 nm. π (0.0200 kg)
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