p39_080 - 80 Letting T e2kL = exp 2L 8 2 m(U E h2 and using...

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80. Letting T e 2 kL =exp à 2 L r 8 π 2 m ( U E ) h 2 ! , and using the result of Exercise 3 in Chapter 39, we solve for E : E = U 1 2 m µ h ln T 4 πL 2 =6
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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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