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1156_Physics ProblemsTechnical Physics

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

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Chapter 41 497 P41.16 (a) x x L x L dx L x x L dx L L = F H G I K J = F H G I K J z z 2 2 2 1 2 1 2 4 2 0 0 sin cos π π x L x L L x L x L x L L L L = + L N M O Q P = 1 2 1 16 4 4 4 2 2 0 2 2 0 π π π π sin cos (b) Probability = F H G I K J = L N M O Q P z 2 2 1 1 4 4 2 0 490 0 510 0 490 0 510 L x L dx L x L L x L L L L L sin sin . . . . π π π Probability = = × 0 020 1 4 2 04 1 96 5 26 10 5 . sin . sin . . π π π a f (c) Probability x L x L L L L N M O Q P = × 1 4 4 3 99 10 0 240 0 260 2 π π 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 n x L dx L 2 2 0 1 sin π F H G I K J = z A n x L dx A L L 2 2 0 2 2 1 sin π F H G I K J = F H G I K J = z or A L = 2 . P41.18 The desired probability is P dx L x L dx L L = = F H G I K J z z ψ π 2 0 4 2 0 4 2 2 sin where sin cos
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