Question

Extra Credit (3 points to Mideterm-2) Q1. A particle is described by the wave function (x) b(a2-x2) for -a sx s a and (x) 0 for x -a and x +a, where a and b are positive real constants. (a) Using the normalization condition, find b in terms a. (b) What is the probability to find the particle at x = +a/2 in a small interval ofwidth 0.01 a ? (c) What is the probability for the particle to be found between x = +a/2 and x = +a ? Q2 A particle is described by the wave function ψ(x) = A x e-bx for 0 the normalization constant A 00, Find x Q3. Suppose the electron is in the 2nd excited state (n-3) in an infinite square potential well. What would be the probability of finding it in a very narrow region Ax 0.01 L wide centered at x = ?

Answer 1

Given Psi (x) = b(a^2 - x^2) for -a lessthanorequalto x lessthanorequalto a = 0 elese where then (a) integral^a_-a Psi* Psi dx = 1 b^2 integral^a_-a (a^2 -x^2)^2 dx = 1 2b^2 integral^a_-0 (a^4 + x^4 - 2a^2 x^2) dx = 1 2b^2[a^4 x + x^5/5 - 2a^2 x^3/3]^a_-0 = 1 2b^2[a^5 + a^5/5 - 2a^5/3] = 1 2b^2 8a^5/15 16/15 a^5 b^2 = 1 so b^2 = 15/16a^5 or b = squareroot 15/16a^5 (b) Probability at x = a/2 in small internal epsilon = 0.01a then integral^x + epsilon_x - epsilon Psi* Psi dx = b^2 integral^x + epsilon_x - epsilon (a^2 - x^2)^2 dx x + epsilon = 0.51a x - epsilon = 0.49a = b^2 integral^0.51a_0.49a (a^2 - x^2)^2 = b^2[a^4 x + x^5/5 - 2a^2 x^3/3]^0.51a_0.49a = b^2[a^4 (0.02a) + 1/5 {(0.510)^5 - (0.49a)^5} - 2/3 a^2{(0.51a - 0.49a}^3] calculate yourself \r\n (c) integral^a_a/2 Psi* Psi dx = b^2 integral^a_a/2 (a^2 - x^2)^2 dx = b^2 integral^a_a/2 (a^4 + x^4 - 2a^2 x^2) dx