1169_Physics ProblemsTechnical Physics

1169_Physics ProblemsTechnical Physics - 510 P41.50 Quantum...

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510 Quantum Mechanics P41.50 (a) The requirement that n L λ 2 = so p hn h L == 2 is still valid. Ep cm c E nhc L mc KEm c nhc L mc mc n nn =+ = F H G I K J + =− = F H G I K J +− bg ej 2 2 2 2 2 2 2 2 2 2 2 2 2 (b) Taking L 100 10 12 . m , m 911 10 31 . kg, and n = 1, we find K 1 14 469 10 J . Nonrelativistic, E h mL 1 2 2 34 2 31 12 2 14 8 6626 10 8 9 11 10 1 00 10 602 10 ×⋅ ×× −− . .. . Js kg m J e j . Comparing this to K 1 , we see that this value is too large by 28 6% . . P41.51 (a) U e d e d ke d e = −+ − +−+ F H G I K J L N M O Q P = 2 0 2 0 2 4 1 1 2 1 3 1 1 2 1 73 4 7 3 ππ af (b) From Equation 41.12, KE h md h e e = 2 2 89 36 1 2 2 2 2 . (c) EUK and dE dd = 0 for a minimum: 7 31 8 0 2 2 2 3 d h e e −= d h kem h mke ee = × × = 3 71 8 42 42 9 11 10 8 99 10 1 60 10 00499 2 2 2 2 34 2 31 9 19 2 a f e j e j . . . C nm .
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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