1156_Physics ProblemsTechnical Physics

# 1156_Physics ProblemsTechnical Physics - Chapter 41 z L...

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Chapter 41 497 P41.16 (a) xx L x L dx L x x L dx LL = F H G I K J =− F H G I K J zz 2 2 21 2 1 2 4 2 00 sin cos ππ x L x L L x L x L x L L L L + L N M O Q P = 1 2 1 16 44 4 2 2 0 2 2 0 π sin cos (b) Probability = F H G I K J L N M O Q P z 2 2 11 4 4 2 0490 0510 L x L dx L x L L x L L L L L sin sin . . . . Probability 0020 1 4 204 196 526 10 5 . sin . sin . . af (c) Probability x L x L L L L N M O Q P 1 4 4 399 10 0240 0260 2 sin . . . (d) In the n = 2 graph in Figure 41.4 (b), it is more probable to find the particle either near x L = 4 or x L = 3 4 than at the center, where the probability density is zero. Nevertheless, the symmetry of the distribution means that the average position is L 2 . P41.17 Normalization requires ψ 2 1 dx all space z = or A nx L dx L 22 0 1 sin F H G I K J = z A L dx A L L 0 2 2 1 sin F H G I K J = F H G I K J = z or A L = 2 . P41.18 The desired probability is Pd x L x L dx == F H G I K J 2 0 4 2 0 4 2 2 sin where sin cos 2 12 2 θ = . Thus, P x L x L L F H G I K J + F H G I K J = 1 4 4 1 4 000
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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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