p40_024 - y L 2 y ! . For n x = n y = 1 , we obtain E 1 , 1...

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24. 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. 40-20 as E nx,ny = 2 h 2 8 m à n 2 x L 2 x + n 2 y L 2 y ! = ( hc ) 2 8( mc 2 ) à n 2 x L 2 x + n 2
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Unformatted text preview: y L 2 y ! . For n x = n y = 1 , we obtain E 1 , 1 = (1240 eV nm) 2 8(511 10 3 eV) 1 (0 . 800 nm) 2 + 1 (1 . 600 nm) 2 = 0 . 73 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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