p40_007 - E = E 4 E 1 = (4 2 1 2 ) h 2 8 m e L 2 = 15( hc )...

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7. We can use the mc 2 value for an electron from Table 38-3 (511 × 10 3 eV) and the hc value developed in problem 3 of Chapter 39 by writing Eq. 40-4 as E n = n 2 h 2 8 mL 2 = n 2 ( hc ) 2 8( mc 2 ) L 2 . The energy to be absorbed is therefore
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Unformatted text preview: E = E 4 E 1 = (4 2 1 2 ) h 2 8 m e L 2 = 15( hc ) 2 8( m e c 2 ) L 2 = 15(1240 eV nm) 2 8(511 10 3 eV)(0 . 250 nm) 2 = 90 . 3 eV ....
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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