1168_Physics ProblemsTechnical Physics

1168_Physics ProblemsTechnical Physics - Chapter 41 P41.47...

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Chapter 41 509 P41.47 xx d x 22 2 = −∞ z ψ For a one-dimensional box of width L , π n L nx L = F H G I K J 2 sin . Thus, x L x L dx LL n L 2 0 2 3 2 = F H G I K J =− z sin (from integral tables). P41.48 (a) 2 1 dx −∞ z = becomes A x L dx A L x L x L A L L L L L 4 4 2 4 4 2 2 2 1 4 4 1 cos sin ππ F H G I K J = F H G I K J + F H G I K J L N M O Q P = F H G I K J L N M O Q P = z or A L 2 4 = and A L = 2 . (b) The probability of finding the particle between 0 and L 8 is 2 0 8 0 8 2 1 4 1 2 0409 dx A x L dx zz = F H G I K J =+ = cos . . P41.49 For a particle with wave function x a e xa a f = 2 for x > 0 and 0 for x < 0. (a) x af 2 0 = , x < 0 and 2 x a e a f = , x > 0 (b) Prob x x dx dx <= = = −∞ −∞ 00 0 2 a f a f a f FIG. P41.49 (c) Normalization ψψ xd x d x d x a f 2 2 0 2 0 1 −∞ −∞ z =+
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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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