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Chapter39

# Chapter39 - 1 Since En L 2 in Eq 39-4 we see that if L is...

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1. 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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2. We first note that since h = 6.626 × 10 –34 J·s and c = 2.998 × 10 8 m/s, hc = ×⋅ × × =⋅ −− 6 626 10 2 998 10 1602 10 10 1240 34 8 19 9 .. . Js m /s J/eV m/nm eV nm. ch c hc h Using the mc 2 value for an electron from Table 37-3 (511 × 10 3 eV), Eq. 39-4 can be rewritten 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 () 222 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 ∆= − = = = = ×
3. We can use the mc 2 value for an electron from Table 37-3 (511 × 10 3 eV) and hc = 1240 eV · nm 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 × = bg ch 8 3 1240 8 511 10 4 7 085 23 eV nm eV eV nm. .

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Alternatively, we can use the mc 2 value for a proton from Table 37-3 (938 × 10 6 eV) and hc = 1240 eV · nm by writing Eq. 39-4 as E nh mL c mc L n p == 22 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. 4. With m = m p = 1.67 × 10 – 27 kg, we obtain () 2 34 2 2 2 21 1 2 2 27 12 6.63 10 J.s 1 3.29 10 J 0.0206eV. 8 8(1.67 10 kg) 100 10 m h En mL ⎛⎞ × ⎜⎟ = × = ⎝⎠ ××
5. 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 1 6.63 10 J 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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6. (a) The ground-state energy is () 2 34 2 2 2 18 1 2 2 31 12 6.63 10 J s 11 . 5 1 1 0 J 8 8(9.11 10 kg) 200 10 m 9.42eV. e h En mL −− ⎛⎞ ×⋅ ⎜⎟ == = × ×× ⎝⎠ = (b) With m p = 1.67 × 10 – 27 kg, we obtain ( ) 2 34 2 2 2 22 1 2 2 27 12 3 6.63 10 J s 1 8.225 10 J 8 8(1.67 10 kg) 200 10 m 5.13 10 eV. p h = ×
7. 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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EE n E E E E E E nn + −= + = = = 11 4 3 2 1 2 21 3 3 4 3 2 1 bg b g c h , we get 2 n + 1 = 21, or n = 10. Thus, (a) the higher quantum number is n + 1 = 10 + 1 = 11, and (b) the lower quantum number is n = 10. (c) Now letting n E E E E E E + + = = = 4 3 2 1 2 2 2 4 3 1 4 b g c h we get 2 n + 1 = 14, which does not have an integer-valued solution. So it is impossible to find the pair of energy levels that fits the requirement. 8. Let the quantum numbers of the pair in question be n and n + 1, respectively. Then E n +1 E n = E 1 ( n + 1) 2 E 1 n 2 = (2 n + 1) E 1 . Letting
9. 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 bg , which leads to 2 n + 1 = 25, or n = 12. Thus, (a) The higher quantum number is n+ 1 = 12+1 = 13.

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Chapter39 - 1 Since En L 2 in Eq 39-4 we see that if L is...

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