ch39-p056 - 56. We can use the mc2 value for an electron...

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(a) With L = 3.0 × 10 9 nm, the energy difference is EE 21 2 39 2 22 1 9 1240 8 511 10 30 10 1 3 1 0 −= ×× −=× ch c h . .e V . (b) Since ( n + 1) 2 n 2 = 2 n + 1, we have EE E h mL n hc mc L n nn =− = + = + + 1 2 2 2 8 8 b g b g c h b g . Setting this equal to 1.0 eV, we solve for n : () ( ) 2 3 9 19 4 4 511 10 eV 3.0 10 nm 1.0eV 11 1.2 10 . 1240eV nm mc L E n hc ∆× × = × (c) At this value of n , the energy is E n = ×≈ × 1240 8 511 10 61 0 0 2 2 18 2 18 c h . eV. Thus, 18 13 23 0eV 1.2 10 . 511 10 eV n E mc × == × × (d) Since
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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