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Unformatted text preview: Chapter 39 1. According to Eq. 394 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 2 2 1 2 , which gives L L 2 . Thus, the ratio is / 2 1.41. L L 2. (a) The groundstate energy is ( ) ( ) ( ) 2 34 2 2 2 18 1 2 2 12 6.63 10 J s 1 1.51 10 J=9.42eV. 8 8 200 10 m e e h E n m L m (b) With m p = 1.67 10 27 kg, we obtain ( ) ( ) ( ) 2 34 2 2 2 22 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 E n m L m 3. To estimate the energy, we use Eq. 394, 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 E n mL 4. With m p = 1.67 10 27 kg, we obtain ( ) ( ) ( ) 2 34 2 2 2 21 1 2 2 12 6.63 10 J.s 1 3.29 10 J 0.0206eV. 8 8 100 10 m p h E n mL m Alternatively, we can use the mc 2 value for a proton from Table 373 (938 10 6 eV) and the hc = 1240 eV nm value developed in problem 83 of Chapter 38 by writing Eq. 394 as 1561 CHAPTER 39 E n h mL n hc m c L n p 2 2 2 2 2 2 2 8 8 bg d i . This alternative approach is perhaps easier to plug into, but it is recommended that both approaches be tried to find which is most convenient. 5. We can use the mc 2 value for an electron from Table 373 (511 10 3 eV) and the hc = 1240 eV nm value developed in problem 83 of Chapter 38 by writing Eq. 394 as E n h mL n hc mc L n 2 2 2 2 2 2 2 8 8 bg c h . For n = 3, we set this expression equal to 4.7 eV and solve for L : L n hc mc E n bg c h b g c hb g 8 3 1240 8 511 10 4 7 085 2 3 eV nm eV eV nm. . . 6. We can use the mc 2 value for an electron from Table 373 (511 10 3 eV) and the hc = 1240 eV nm value developed in problem 83 of Chapter 38 by writing Eq. 394 as E n h mL n hc mc L n 2 2 2 2 2 2 2 8 8 bg c h . The energy to be absorbed is therefore ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 4 1 2 2 2 2 3 4 1 15 15 1240eV nm 90.3eV. 8 8 8 511 10 eV 0.250nm e e h hc E E E m L m c L D 7. Since E n L 2 in Eq. 394, we see that if L is doubled, then E 1 becomes (2.6 eV)(2) 2 = 0.65 eV. 8. Let the quantum numbers of the pair in question be n and n + 1, respectively. We note that E E n h mL n h mL n h mL n n 1 2 2 2 2 2 2 2 2 1 8 8 2 1 8 b g b g Therefore, E n +1 E n = (2 n + 1) E 1 . Now E E E E E n E n n 1 5 2 1 1 1 5 25 2 1 b g , 26 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....
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This note was uploaded on 04/19/2008 for the course PHYS 1100, 1200 taught by Professor Many during the Fall '08 term at Rensselaer Polytechnic Institute.
 Fall '08
 Many
 Physics, Energy

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