1166_Physics ProblemsTechnical Physics

1166_Physics ProblemsTechnical Physics - Chapter 41 P41.43...

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Chapter 41 507 P41.43 R k k k k k k k k = + = + 1 2 2 1 2 2 2 1 2 2 1 2 1 1 b g b g b g b g = 2 2 2 k m E U = for constant U = 2 1 2 2 k m E = since U = 0 (1) = 2 2 2 2 k m E U = (2) Dividing (2) by (1), k k U E 2 2 1 2 1 1 1 2 1 2 = = = so k k 2 1 1 2 = and therefore, R = + = + = 1 1 2 1 1 2 2 1 2 1 0 029 4 2 2 2 2 e j e j e j e j . . FIG. P41.43 P41.44 (a) The wave functions and probability densities are the same as those shown in the two lower curves in Figure 41.4 of the textbook. (b) P dx x dx x x 1 1 2 0 150 2 0 150 0 350 0 150 2 1 00 1 00 2 00 2 1 00 2 1 00 = = F H G I K J F H G I K J = F H G I K J L N M O Q P z z ψ π π π . . . . . sin . . . sin . nm 0.350 nm nm 0.350 nm nm nm nm nm 4 nm b g In the above result we used sin sin 2 2 1 4 2 axdx x a ax z = F H G I K J F H G I K J a f . Therefore, P x x 1 0 150 1 00 1 00 2 1 00 = F H G I K J L N M O Q P . . sin . . nm nm 2 nm nm 0.350 nm b g π π P 1 1 00 0 350 0 150 1 00 0 700 0 300 0 200 = R S T U V W = . . . . sin . sin . . nm nm nm nm 2 b g a f a f π π π . (c) P x dx x x 2 2
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