p40_025 - = n y = n z = 1 , we obtain E 1 , 1 = (1240 eV...

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25. 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. 40-21 as E nx,ny,nz = 2 h 2 8 m à n 2 x L 2 x + n 2 y L 2 y + n 2 z L 2 z ! = ( hc ) 2 8( mc 2 ) à n 2 x L 2 x + n 2 y L 2 y + n 2 z L 2 z ! . For n x
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Unformatted text preview: = n y = n z = 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 + 1 (0 . 400 nm) 2 = 3 . 1 eV ....
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