1166_Physics ProblemsTechnical Physics

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

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Chapter 41 507 P41.43 R kk = + = + 12 2 2 21 2 2 1 1 bg = 22 2 k m EU =− for constant U = 2 1 2 2 k m E = since U = 0 (1) = 2 2 2 2 k m (2) Dividing (2) by (1), k k U E 2 2 1 2 11 1 2 1 2 =− =− = so k k 2 1 1 2 = and therefore, R = + = + = 11 2 00294 2 2 2 2 ej .. 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) Pd x x dx x x 2 0150 2 0350 2 100 200 2 2 == F H G I K J F H G I K J F H G I K J L N M O Q P zz ψ π . . . sin . . . sin . nm 0.350 nm nm 0.350 nm nm nm nm nm 4 n m 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 af . Therefore, Px x 1 2 F H G I K J L N M O Q P . . sin . . nm nm 2 n m nm 0.350 nm P 1 1 00 0 350 0 150 0 700 0 300 0 200 −− R S T U V W = . . sin . sin . . nm nm nm nm 2 ππ . (c) P x dx x x 2 2 2 2 2 8 4 = F H G I K J F H G I K J L N M O Q P z . sin . . . sin . . . . . x 2 4 4 1 00 0 350 0 150 4 140 0600 0351
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