Physics exercises_solution: Chapter 42

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Physics exercises_solution: Chapter 42

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Bộ tài liệu bài tập tham khảo môn vật lý bậc đại học bằng tiếng anh Chapter 42

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Nội dung Text: Physics exercises_solution: Chapter 42

  1. 3 2 K 2(7.9  10 4 eV)(1.60  10 19 J eV) 42.1: a) K  kT  T   2 3k 3(1.38  10 23 J K)  T  6.1 K 2(4.48 eV) (1.60  1019 J eV) b) T   34,600 K. 3(1.38  10 23 J K ) c) The thermal energy associated with room temperature (300 K) is much greater than the bond energy of He 2 (calculated in part (a)), so the typical collision at room temperature will be more than enough to break up He 2 . However, the thermal energy at 300 K is much less than the bond energy of H 2 , so we would expect it to remain intact at room temperature. 1 e2 42.2: a) U    5.0 eV. 4πε0 r b)  5.0 eV  (4.3 eV  3.5 eV)  4.2 eV. 42.3: Let 1 refer to C and 2 to O. m1  1.993  10 26 kg, m2  2.656  10 26 kg, r0  0.1128 nm  m2  r1    m  m r0  0.0644 nm (carbon )   1 2   m1  r2    m  m r0  0.0484 nm (carbon)   1 2  b) I  m1r12  m2 r22  1.45  10 46 kg  m 2 ; yes, this agrees with Example 42.2. 42.4: The energy of the emitted photon is 1.01  10 5 eV, and so its frequency and wavelength are E (1.01  105 eV)(1.60  1019 J eV) f    2.44 GHz h (6.63  1034 J  s) c (3.00  108 m s) λ   0.123 m. f (2.44  109 Hz) This frequency corresponds to that given for a microwave oven.
  2. 42.5: a) From Example 42.2, E1  0.479 meV  7.674  10 23 J and I  1.449  10 46 kg  m 2 K  1 Iω 2 and K  E gives ω  2 E1 I  1.03  1012 rad s 2 b) v1  r1ω1  (0.0644  109 m)(1.03  1012 rad s)  66.3 m s (carbon) v2  r2 ω2  (0.0484  10 9 m)(1.03  1012 rad s)  49.8 m s (oxygen) c) T  2π ω  6.10  10 12 s 1 1 42.6: a) E0  ω  (0.2690 eV)  2.083  10  20 J 2 2 1 2 E0 2(2.083  10 2 J) E0  mr vmax gives vmax  2   26  1.91  103 m s 2 mr 1.139  10 kg b) According the Eq. 42.7 the spacing between adjacent vibrational energy levels is twice the ground state energy: 1 En  (n  ) ω, 2 E  En 1  En  ω  hf . Thus, using the E specified in Example 42.3, it follows that its vibrational period is 1 h (6.63  1034 J  s) T    1.54  1014 s. f E (0.2690 eV)(1.60  1019 J eV) c) The vibrational period is shorter than the rotational period.
  3.  m m  42.7: a) I  mr r 2   Li H  r 2 m m   Li H  (1.17  10 26 kg)(1.67  10 27 kg)(1.59  10 10 m) 2  (1.17  10 26 kg  1.67  10 27 kg)  3.69  10 47 kg  m 2 2 42 E  E4  E3  (4(4  1)  (3) (3  1))  2I I 34 4(1.054  10 J  s) 2  E  3.69  10 47 kg  m 2  E  1.20  10 21 J  7.53  10 3 eV. hc (6.63  1034 J  s)(3.00  108 m s) b) λ    21  λ  1.66  10 4 m. E 1.20  10 J 42.8: Each atom has a mass m and is at a distance L 2 from the center, so the moment of inertia is 2(m)( L 2) 2  mL2 2  2.21  10  44 kg  m 2 . l (l  1)2 l (l  1)2 2 2 l2 42.9: a) El  , El 1   E  (l  l  l 2  l )  2I 2I 2I I ΔE ΔE l b) f    . h 2π 2πI hc 42.10: a) E    k  mr , and solving for k , λ 2  2πc  k    mr  205 N m.  λ 
  4.  1 2 42.11: Energy levels are E  En  El   n   ω  l (l  1)  2 2I  1   n   (0.269 eV)  l (l  1) (2.395  10 4 eV) where the values are from  2 Examples 42.2 and 42.3. a) n  0, l  1  n  1, l  2 :  E  E f  Ei  (1)(0.2690 eV)  (4)(2.395  104 eV)  0.2700 eV hc (6.626  1034 J  s)(2.998  108 m s) λ  E (0.2700 eV)(1.602  1019 J eV)  4.592  10 6 m. b) n  0, l  2  n  1, l  1 :  E  E f  Ei  (1)(0.2690 eV)  (4)(2.395  104 eV)  0.2680 eV hc (6.626  10 34 J  s)(2.998  108 m s) λ  E (0.2680 eV)(1.602  1019 J eV)  λ  4.627  10 6 m c) n  0, l  3  n  1, l  2 : E  E f  Ei  (1)(0.2690 eV)  (6)(2.395  104 eV)  0.2676 eV hc (6.626  1034 J  s)(2.998  108 m s) λ  19  4.634  10 6 m. ΔE (0.2676 eV)(1.602  10 J eV) 42.12: ω   k  mr   2k  m  3.14  10 20 J  0.196 eV, where mr  m 2 has been used.
  5. ω 1 k 42.13: a) f   2π 2π mr m1m2 2πf  2  k   mr 2πf   2 m1  m2 (1.67  10 27 kg ) (3.15  10 26 kg)[2π (1.24  1014 Hz)]2  k  (1.67  10 27 kg  3.15  10 26 kg)  k   963 N m  3  1 b) E   n   ω   n   ω  ω  hf  2  2    E  6.63  1034 J  s 1.24  1014 Hz  8.22  1020 J   0.513 eV c 3.00  108 m s c) λ    2.42  10 6 m infrared  f 1.24  10 Hz 14 42.14: a) As a photon, λ hc    6.63  1034 J  s 3.00  108 m s  0.200 nm.  E   6.20  103 eV 1.60  1019 J eV  b) As a matter wave, λ h  h  6.63  10 J  s 34  0.200 nm and p 2mE  2 9.11  10 31 kg  37.6 eV  1.60  10 19 J eV  c) as a matter wave, λ h   6.63  1034 J  s   0.200 nm . 2mE   2 1.67  10 27 kg 0.0205 eV  1.60  1019 J eV   42.15: The volume enclosing a single sodium and chlorine atom  2 2.82  1010 m    3 4.49  1029 m 3 . So the density m  mCl 3.82  1026 kg  5.89  1026 kg ρ  Na  V 4.49  10 29 m3  ρ  2.16  103 kg m3 .
  6. 42.16: For an average spacing a, the density is ρ  m a 3 , where m is the average of the ionic masses, and so a   3  m 6.49  10 26 kg  1.33  10 25 kg 2   3.60  10 29 m 3 , ρ (2.75  10 kg m ) 3 3 and a  3.30  10 10 m  0.330 nm . b) The larger (higher atomic number) atoms have the larger spacing. hc (6.63  10 34 J  s) (3.00  108 m s) 42.17: E   13  2.14  10 13 J λ 9.31  10 m  1.34  106 eV 1.34  106 eV So the number of electrons that can be excited to the conduction band is n  1.12 eV  1.20  106 electrons. hc 42.18: a)  2.27  10 7 m  227 nm , in the ultraviolet. E b) Visible light lacks enough energy to excite the electrons into the conduction band, so visible light passes through the diamond unabsorbed. c) Impurities can lower the gap energy making it easier for the material to absorb shorter wavelength visible light. This allows longer wavelength visible light to pass through, giving the diamond color. 42.19: a) To be detected the photon must have enough energy to bridge the gap width E  1.12 eV λ hc    6.63  1034 J  s 3.00  108 m s   1.11  10 6 m , in the infrared. E  1.12 eV  1.60  10 J eV 19  b) Visible photons have more than enough energy to excite electrons from the valence to conduction band. Thus visible light is absorbed, making silicon opaque. 42.20: vrms  3kT m  1.17  105 m s , as found in Example 42.9. The equipartition theorem does not hold for the electrons at the Fermi energy. Although these electrons are very energetic, they cannot lose energy, unlike electrons in a free electron gas.
  7.  2   2ψ  2ψ  2ψ  42.21: a)  2  2  2   Eψ  x 2m  y z   nx πx n πy n πz where ψ  A sin sin y sin z L L L 2  2ψ n π  2ψ  2ψ    x  ψ , and similarly for and 2 . x 2  L  y 2 z 2   n x π   n y π   n z π   2 22           ψ  Eψ 2m  L   L   L       E  nx  n y  nz π  2 2 2 2 2  2mL2 3π 2 2 b) Ground state  nx  n y  nz  1  E  , 2mL2 The only degeneracy is from the two spin states. The first excited state  2, 1, 1 or 1, 2, 1 3π 2 2 or 1, 1, 2  E  and the degeneracy is 2  3  6 . mL2 9π 2 2 The second excited state  2, 2, 1 or 2, 1, 2 or 1, 2, 2  E  and the 2mL2 degeneracy is 2   3  6 . 1   ψ dV 2 42.22:  L 2  nx πx    L 2  n y πy    L 2  nz πz    A   sin  2   dx    sin   L  dy    sin  L  dz    0  L  0    0     3  L A  , 2 2 so A  2 L  (assuming A to be real positive). 32
  8. 42.23: Density of states: g E   2m 3 2 V E 1 2 2π 2 3 (2(9.11  10 31 kg)) 3 2 (1.0  10 6 m 3 )(5.0 eV)1 2 (1.60  10 19 J eV)1 2  g E   2π 2 (1.054  10 34 J  s) 3    9.5  10 40 states J 1.60  10 19 J eV   1.5  10 states eV. 22 42.24: Equation (42.13) may be solved for nrs  2mE  L π  , and substituting this 12 into Eq. (42.12), using L3  V , gives Eq. (42.14). n 2 π 2 2 42.25: Eq.(42.13): E  rs 2 2mL L  nrs  2mE π 0.010 m  2(9.11  10 31 kg) (7.0 eV) (1.60  10 19 J eV) π (1.054  10 34 J  s)  nrs  4.3  10 7. 3 42.26: a) From Eq. (42.22), Eav  EF  1.94 eV. 5 b) 2E m   21.94 1.60  10 19 J eV   8.25  105 m s . 9.11  10 31 kg c)   EF 3.23 eV  1.60  1019 J eV   3.74  104 K. k  1.38  10 23 J K   π 2 kT   π 2 1.38  10 23 J K  300 K   42.27: a) CV    2E  R  25.48 eV  1.60  1019 J eV   R   F     CV  0.0233 R  0.194 J mol K. 0.194 J mol K b)  7.68  10 3 . 25.3 J mol K c) Mostly ions (see Section 18.4).
  9. 42.28: a) See Example 42.10: The probabilities are 1.78  10 7 , 2.37  10 6 , and 1.51 105 . b) The Fermi distribution, Eq. (42.17), has the property that f EF  E   1  f E  (see Problem (42.46)), and so the probability that a state at the top of the valence band is occupied is the same as the probability that a state of the bottom of the conduction band is filled (this result depends on having the Fermi energy in the middle of the gap). 1 42.29: f E    E  EF  kT e 1  1   EF  E  kT ln   f E   1      EF  E  1.38  10 23 J K 300 K  ln  1 4   1  4.4  10    EF  E  3.20  10  20 J  E  0.20 eV. So the Fermi level is 0.20 eV below the conduction band. 42.30: a) Solving Eq. (42.23) for the voltage as a function of current, kT  I  kT  40.0 mA  V ln   1  I  e ln  3.60 mA  1  0.0645 V.   e  S    b) From part (a), the quantity eeV kT  12.11 , so far a reverse-bias voltage of the same magnitude,   1 I  I S e  eV kT  1  I S     1  3.30 mA .  12.11 
  10. 42.31:  I  IS eeV kT   1  IS  e 1 eV kT 1 a)   eV 1.60  1019 C 1.50  102 V   0.580  kT  1.38  10 23 J K 300 K   9.25  103 A  IS   0.0118 A. e0.580  1 Now for V  0.0100 V, eV kT  0.387    I  0.0118 A  e 0.387  1  5.56  10 3 A  5.56 mA eV b) Now with V  15.0 mV,  0.580 kT    I  0.0118 A  e 0.580  1  5.18  103 A . eV If V  10.0 mV   0.387 kT    I  0.0118 A  e 0.387  1  3.77  10 3 A . 22 hλ 42.32: See Problem (42.7): I   2  7.14  10 48 kg  m 2 . E 2π c 42.33:    a) p  qd  1.60  10 19 C 2.4  10 10 m  3.8  10 29 C  m p 3.0  10 29 C  m b) q   10  1.3  10 19 C d 2.4  10 m c) q e  0.78 p 1.5  1030 C  m q  10  9.4  10 21 C d) d 1.6  10 m  q e  0.059 This is much less than for sodium chloride (part (c)). Therefore the bond for hydrogen iodide is more covalent in nature than ionic.
  11. 42.34: The electrical potential energy is U  5.13 eV, and 1 e2 r  2.8  10 10 m. 4πε0 U 42.35: a) For maximum separation of Na  and Cl  for stability:  e2 U 4πε0 r    5.1 eV  3.6 eV  1.60  10 19 J eV  2.40  10 19 J (1.60  10 19 C) 2 r 19  9.6  10 10 m. 4πε0 (2.40  10 J ) b) For K  and Br  : U  e2 4πε0 r    4.3 eV  3.5 eV  1.60  10 19 J eV  1.28  10 19 J (1.60  10 19 C) 2 r  1.8  10 9 m. 4πε0 (1.28  10 19 J) 42.36: The energies corresponding to the observed wavelengths are 3.29  1021 J, 2.87  10 21J, 2.47  10 21 J, 2.06  10 21 J and 1.65  10 21 J. The average spacing of these energies is 0.410  10 21 J and using the result of Problem (44-4), these are seen to correspond to transition from levels 8, 7, 6, 5 and 4 to the respective next lower levels. 2 Then,  0.410  10 21 J , from which I  2.71  10 47 kg  m 2 . I
  12. I I (mH  mCl ) 42.37: a) Pr. (44.36) yields I  2.71  10 47 kg  m 2 , and so r   mr mH mCl (2.71  10 47 kg  m 2 ) (1.67  10 27 kg  5.81  10 26 kg) r (1.67  10 27 kg  5.81  10 26 kg)  r  1.29  10 10 m. 2 l2 b) From l  l  1 : E  l l  1  l  1l   . But E  hc  l  2πcI  2I I λ λ 4.84  10 4 m . So the l-values that lead to the wavelength of Pr. (44-32) are: λ 4.84  10 4 m λ  6.04  10 5 m : l   8. 6.04  10 5 m Similarly for: λ  6.90  10 5 m : l  7; λ  8.04  10 5 m : l  6 λ  9.64  10 5 m : l  5; λ  1.204  10  4 m : l  4. c) The longest wavelength means the least transition energy (l  1  l  0) (1) (1.054  10 34 J  s) 2  E   47  4.10  10 22 J 2.71  10 kg  m 2 hc λ  4.85  10 4 m. E d) If the hydrogen atom is replaced by deuterium, then the reduced mass changes to mr  3.16  1027 kg. Now, l2 hc 2πcI  2πcmr r 2 E    λ   I λ l l  m   3.16  10 kg   27  λ   r λ   m   1.62  10 27 kg λ  (1.95)λ   r   So for l 8l 7: λ  (60.4 μm) (1.95)  118 μm. l  7 l  6: λ  (69.0 μm) (1.95)  134 μm. l  6  l  5: λ  (80.4 μm) (1.95)  156 μm. l  5l  4: λ  (96.4 μm) (1.95)  188 μm. l  4  l  3: λ  (120.4 μm) (1.95)  234 μm.
  13. 42.38: From the result of Problem (42.9), the moment inertia of the molecule is 2l hlλ I  2  6.43  10 46 kg  m 2 E 4π c and from Eq. (42.6) the separation is I r0   0.193 nm. mr
  14. L2 2l (l  1) 42.39: a) Eex   . 2I 2I E g  0 (l  0), and there is an additional multiplicative factor of 2l + 1 because for each l state there are really 2l  1 ml states with the same energy.   l ( l 1)  2 nl   kT So  (2l  1)e   2 I     n0 b) T  300 K, I  1.449  10 46 kg  m 2 . 2 (1) (1  1) (i) El 1   46  7.67  10 23 J. 2(1.449  10 kg  m ) 2 El 1 7.67  10 23 J   0.0185. kT (1.38  10 23 J K) (300 K) (2l  1)  3 n so l  1  (3)e  0.0185  2.95. n0 El 2 2 (2) (2  1) (ii)   0.0556. kT 2(1.449  10 46 kg  m 2 ) (1.38  10 23 J K ) (300 K) (2l  1)  5. nl 1  (5)(e 0.0556 )  4.73. n0 El 10 2 (10) (10  1) (iii)   1.02. kT 2(1.449  10 46 kg  m 2 ) (1.38  10 23 J K ) (300 K) n (2l  1)  21 so l 10  (21) (e 1.02 )  7.57. n0 El 20 2 (20) (20  1) (iv)   3.89. kT 2(1.449  10 46 kg  m 2 ) (1.38  10 23 J K ) (300 K) (2l  1)  41. nl 20  (41)e 3.89  0.838. n0
  15. 42.40: a) I co  1.449  10 46 kg  m 2 . 2l (l  1) 1.054  1034 J  s) 2 (1) (1  1) El 1    7.67  10 23 J. 2I 2(1.449  10 46 kg  m 2 ) El 0  0. E  7.67  10 23 J  4.79  10 4 eV. hc (6.63  10 34 J  s) (3.00  108 m s) λ   2.59  10 3 m  2.59 mm. E (7.67  10 23 J) b) Let’s compare the value of kT when T=20 K to that of E for the l  1  l  0 rotational transition: kT  (1.38  10 23 J K) (20 K)  2.76  10 22 J. kT E  7.67  10 23 J (from part (a)). So  3.60. E Therefore, although T is quite small, there is still plenty of energy to excite CO molecules into the first rotational level. This allows astronomers to detect the 2.59 mm wavelength radiation from such molecular clouds.
  16. mNa mCl 2 42.41: a) I  mr r 2  r mNa  mCl (3.8176  10 26 kg) (5.8068  10 26 kg) (2.361  10 10 m) 2  (3.8176  10 26 kg  5.8068  10 26 kg)  I  1.284  10 45 kg  m 2 For l  2  l  1 : 2 2(1.054  10 34 J  s) 2  E  E2  E1  (6  2)   1.730  10  23 J 2 I 1.284  10  45 kg  m 2 hc λ  0.01148 m  1.148 cm. E 2 1 For l  1  l  0 : E  E1  E0  (2  0)  (1.730  10 23 J) 2I 2  8.650  10 24 J hc λ  2.297 cm. E b) Carrying out exactly the same calculation for Na 37 Cl, where mr (37)  2.354  10 26 kg and I(37)  1.312  10 45 kg  m 2 we find for l  2  l  1: E  1.693  10 23 J and λ  1.173 cm. For l  1  l  0 : E  8.465  10  24 J and λ  2.347 cm. So the differences in wavelength are: l  2  l  1 : λ  1.173 cm  1.148 cm  0.025 cm. l 1 l  0: λ  2.347 cm  2.297 cm  0.050 cm. E 42.42: The vibration frequency is, from Eq. (42.8), f   1.12  1014 Hz. The force h constant is k   (2πf ) 2 mr  777 N m .  1 k 1 2k  42.43: En   n     E0    2 mr 2 mH 1 2(576 N m)  E0  (1.054  10 34 J  s)  4.38  10 20 J  0.274 eV. 2 1.67  10 27 kg This is much less than the H 2 bond energy.
  17. 42.44: a) The frequency is proportional to the reciprocal of the square root of the reduced mass, and in terms of the atomic masses, the frequency of the isotope with the deuterium atom is 12 12  m m (mH  mF )   1  (mF mD )  f  f0  F H  m m ( m  m )   f 0  1  (m m )  .     F D D F   F H  Using f 0 from Exercise (42.13) and the given masses, f  8.99  1013 Hz. mH mI r 2 42.45: a) I  mr r 2   (1.657  10 27 kg) (0.160  10 9 m) 2 mH  mI  4.24  10 47 kg  m 2 b) Vibration-rotation energy levels are: 2  1 k El  l (l  1)  h    2I  2 mr 2  1  k   l (l  1)   n   hf  since : ω  2πf     2I  2  mr  i) n  1, l  1  n  0, l  0 : 2  3 1  2 E  (2  0)     hf   hf . 2I  2 2  I hc hc c 3.00  108 m s λ  2   . E   3.96  1011 Hz  6.93  1013 μ  hf  f I 2I  λ  4.30  10 6 m. ii) n  1, l  2  n  0, l  1 : 2 E  (6  2)  hf 2I c 3.00  108 m s λ   4.28  10 6 m.  2  2(3.96  1011 Hz)  6.93  1013 Hz 2  2πI   f    iii) n  2, l  2  n  1, l  3 2 E  (6  12)  hf 2I c 3.00  108 m s λ   4.40  10 6 m.    2  3(3.96  1011 Hz)  6.93  1013 Hz  3 2I   f   
  18. 42.46: The sum of the probabilities is 1 1 f ( EF  E )  f ( EF  E )   E kT  E kT e 1 e 1  E / kT 1 e   E kT  e  1 1  e E kT  1. 42.47: Since potassium is a metal we approximate E F  E F0 . 32 3 π 4 3 2 n 2 3  EF  2m ρ But the electron concentration n  m 851 kg m 3 n  26  1.31  10 28 electron m 3 6.49  10 kg 32 3 π 4 3 (1.054  10 34 J  s) 2 (1.31  10 28 /m 3 ) 2 3  EF  31  3.24  10 19 J  2.03 eV. 2(9.11  10 kg) 42.48: a) First we calculate the number-density of neutrons from the given mass-density: n  (7.0  1017 kg m 3 ) / 1.67  10 27 kg neutron )  4.2  10 44 m 3 . Now use Eq. 44.21 2 4 1 2 4 2 3 3 π 3 2 n 3 3 3 π 3 (6.63  10 34 J  s 2π ) 2 (4.2  10 44 m 3 ) 3 EF 0    27  1.8  10 11 J. 2m 2(1.67  10 kg ) b) Set kT  EF0 (see Exercise 42.26) to obtain EF 0 (1.8  10 11 J ) T   1.3  1012 K. k (1.38  10 28 J K ) 42.49: a) Each unit cell has one atom at its center and 8 atoms at its corners that are each shared by 8 other unit cells. So there are 1  8 8  2 atoms per unit cell. n 2  9  4.66  10 28 atoms m 3 V (0.35  10 m) 3 23 32 3 π 4 3 2  N  b) E F0    2m  V  In this equation N V is the number of free electrons per m 3 . But the problem says to assume one free electron per atom, so this is the same as n V calculated in part (a). m  9.109  10 31 kg (the electron mass), so EF0  7.563  10 19 J  4.7 eV
  19. d αe 2 1 1 42.50: a) U tot  2  8A 9 . dr 4πε0 r r Setting this equal to zero when r  r0 gives 8 A4πε0 r07  αe 2 and so αe 2  1 r07  U tot    . 4πε0  r 8r 8    At r  r0 , 7αe 2 U tot    1.26  10 18 J  7.85 eV. 32πε0 r0 b) To remove a Na  Cl  ion pair from the crystal requires 7.85 eV. When neutral Na and Cl atoms are formed from the Na  and Cl  atoms there is a net release of energy  5.14 eV  3.61 eV  1.53 eV, so the net energy required to remove a neutral Na, Cl pair from the crystal is 7.85 eV  1.53 eV  6.32 eV. 23 3 3 32 3 π 4 3 2  N  42.51: a) Eav  EF 0    . Etot  NEav . 5 5 2m  V  1 3 dEtot 2 3 32 3 π 4 3 2  N    N  N       dV 3 5 2m  V   V 2  53 32 3 π 4 3 2 N    . 5m V  53 dEtot 32 3 π 4 3 2  N  P    . dV 5m  V  N b)  8.45  10 28 m 3 V 32 3 π 4 3 (1.054  10 34 J  s) 2  p (8.45  10 28 m 3 ) 5 3 5(9.11  10 31 kg)  3.80  1010 Pa  3.76  105 atm (!) c) There is a large attractive force on the electrons by the copper ions.
  20. 42.52: a) From Problem (42.51): 53 32 3 π 4 3 2  N  p   . 5m  V  dp  5 32 3 π 4 3 2  N  2 3   N  B  V  V       2  dV 3  5m  V   V   5  p. 3 N b)  8.45  10 28 m 3 . V 5 32 3  4 32 B  (8.45  10 28 m 3 ) 5 3  6.33  1010 Pa. 3 5m 6.33  1010 Pa c)  0.45. The copper ions themselves make up the remaining 1.4  1011 Pa fraction. 23 32 3 π 4 3 2 N 42.53: a) EF0    . 2m V  32 1 N  2m 2 c 2  23 2 m 3c 3 23 2 m 3c 3 Let EF 0  mc 2      23 43 2    100  V   (100)3 π   1003 23π 2 3 3000π 2 3 3 1.67  10 m . 33 8.45  10 28 m 3 b)  5.06  10 5. 1.67  1033 m 3 Since the real concentration of electrons in copper is less than one part in 10 4 of the concentration where relativistic effects are important, it is safe to ignore relativistic effects for most applications. 6(2  1030 kg) c) The number of electrons is N e   6.03  1056. The 1.99  10 26 kg Ne 6.03  1056 concentration is  4  6.66  1035 m 3 . V 3 π (6.00  106 m) 3 6.66  1035 m 3 d) Comparing this to the result from part (a)  400 so relativistic effects 1.67  1033 m 3 will be very important.
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