1170_Physics ProblemsTechnical Physics

1170_Physics ProblemsTechnical Physics - Chapter 41 P41.52...

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Chapter 41 511 P41.52 (a) ψ ω = Bxe mx 2 2 = bg d dx Be Bx m xe Be B m xe d dx Bx m xe B m xe B m x m xe d dx B m xe m ψω ωω =+ F H G I K J =− F H G I K J F H G I K J F H G I K J F H G I K J F H G I K J F H G I K J −− 22 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3 == = = = = = = = xm x B m 2 3 2 + F H G I K J = = Substituting into the Schrödinger Equation (41.13), we have F H G I K J + F H G I K J + F H G I K J 3 2 2 2 3 2 2 2 2 2 2 2 2 B m xe B m mE Bxe m xB x e = = = = . This is true if −= 3 2 E = ; it is true if E = 3 2 = . (b) We never find the particle at x = 0 because = 0 there. (c) is maximized if d dx x m ==− F H G I K J 01 2 = , which is true at x m = . (d) We require 2 1 dx −∞ z = : 12 2 1 42 2 2 2 3 21 2 3 2 32 = = −∞ zz Bxe d x B xe d x B m B m π = = af . Then B mm = F H G I K J = F H G I K J 24 14 34 33 3 = = . (e) At x m = 2 = , the potential energy is 1 2 1 2 4 2 2 m m = F H G I K J = = = . This is larger than the total energy 3 2 = , so there is zero classical probability of finding the particle here. (f) Probability F H I K = ψδ δ 2 2 2 dx
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