ch39 - 1. According to Eq. 39-4 En L 2. As a consequence,...

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1. According to Eq. 39-4 E n L – 2 . As a consequence, the new energy level E' n satisfies = F H G I K J = F H G I K J = E E L L L L n n 22 1 2 , which gives ′ = LL 2 . Thus, the ratio is / 2 1.41. ==
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2. (a) The ground-state energy is () 2 34 2 2 21 8 1 2 2 12 6.63 10 J s 1 1.51 10 J=9.42eV. 8 8 200 10 m e e h En mL m §· ×⋅ ¨¸ == = × × ©¹ (b) With m p = 1.67 × 10 – 27 kg, we obtain 2 34 2 2 22 2 3 1 2 2 12 6.63 10 J s 1 8.225 10 J=5.13 10 eV. 8 8 200 10 m p p h m −− = × × ×
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3 . To estimate the energy, we use Eq. 39-4, with n = 1, L equal to the atomic diameter, and m equal to the mass of an electron: () ( ) 2 2 34 2 2 10 2 2 31 14 16 . 6 3 1 0J s 3.07 10 J=1920MeV 1.9 GeV. 8 8 9.11 10 kg 1.4 10 m h En mL −− ×⋅ == = × ××
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4. With m p = 1.67 × 10 – 27 kg, we obtain () 2 34 2 2 22 1 1 2 2 12 6.63 10 J.s 1 3.29 10 J 0.0206eV. 8 8 100 10 m p h En mL m §· × ¨¸ == = × = ©¹ × Alternatively, we can use the mc 2 value for a proton from Table 37-3 (938 × 10 6 eV) and the hc = 1240 eV · nm value developed in problem 83 of Chapter 38 by writing Eq. 39-4 as E nh mL c mc L n p 2 2 2 8 8 b g di . This alternative approach is perhaps easier to plug into, but it is recommended that both approaches be tried to find which is most convenient.
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5. We can use the mc 2 value for an electron from Table 37-3 (511 × 10 3 eV) and the hc = 1240 eV · nm value developed in problem 83 of Chapter 38 by writing Eq. 39-4 as E nh mL c mc L n == 22 2 2 2 8 8 b g c h . For n = 3, we set this expression equal to 4.7 eV and solve for L : L nhc mc E n × = b g c h b g c hb g 8 3 1240 8 511 10 4 7 085 23 eV nm eV eV nm. .
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6. We can use the mc 2 value for an electron from Table 37-3 (511 × 10 3 eV) and the hc = 1240 eV · nm value developed in problem 83 of Chapter 38 by writing Eq. 39-4 as E nh mL c mc L n == 22 2 2 2 8 8 b g c h . The energy to be absorbed is therefore () 2 41 2 2 3 15 15 1240eV nm 90.3eV. 8 8 8 511 10 eV 0.250nm e e h hc EE E mL mc L ∆= − = = = = ×
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7. Since E n L – 2 in Eq. 39-4, we see that if L is doubled, then E 1 becomes (2.6 eV)(2) – 2 = 0.65 eV.
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8. Let the quantum numbers of the pair in question be n and n + 1, respectively. We note that EE nh mL mL mL nn + −= + + 1 2 2 2 22 2 2 2 1 88 21 8 b g b g Therefore, E n +1 E n = (2 n + 1) E 1 . Now E E E n E + −== = = + 15 2 11 1 52 1 b g , which leads to 2 n + 1 = 25, or n = 12. Thus, (a) the higher quantum number is n+ 1 = 12+1 = 13, and (b) the lower quantum number is n = 12. (c) Now let E E E n E + = = + 16 2 1 63 62 1 b g , which gives 2 n + 1 = 36, or n = 17.5. This is not an integer, so it is impossible to find the pair that fits the requirement.
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9. The energy levels are given by E n = n 2 h 2 /8 mL 2 , where h is the Planck constant, m is the mass of an electron, and L is the width of the well. The frequency of the light that will excite the electron from the state with quantum number n i to the state with quantum number n f is fE h h m L n n fi == 8 22 2 c h di and the wavelength of the light is λ= = c f mL c hn n 8 2 .
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ch39 - 1. According to Eq. 39-4 En L 2. As a consequence,...

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