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Unformatted text preview: 12. We can use the mc 2 value for an electron from Table 383 (511 10 3 eV) and the hc value developed in problem 3 of Chapter 39 by rewriting Eq. 404 as E n = n 2 h 2 8 mL 2 = n 2 ( hc ) 2 8( mc 2 ) L 2 . (a) The first excited state is characterized by n = 2, and the third by n = 4. Thus, E = ( hc ) 2 8( mc 2 ) L 2 n 2 n 2 = (1240 eV nm) 2 8(511 10 3 eV)(0 . 250 nm) 2 ( 4 2 2 2 ) = (6 . 02 eV)(16 4) which yields E = 72 . 2 eV . (b) Now that the electron is in the n = 4 level, it can drop to a lower level ( n ) in a variety of ways. Each of these drops is presumed to cause a photon to be emitted of wavelength = hc E n E n = 8( mc 2 ) L 2 hc ( n 2 n 2 ) . For example, for the transition n = 4 to n = 3, the photon emitted would have wavelength = 8(511 10 3 eV)(0 . 250 nm) 2 (1240 eV nm) (4 2 3 2 ) = 29 . 4 nm , and once it is then in level n = 3 it might fall to level n = 2 emitting another photon. Calculating in this way all the possible photons emitted during the deexcitation of this system, we find...
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 Fall '08
 SPRUNGER
 Physics

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