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Chapter 35: Problem 11

A particle's wave function is \(\psi=A e^{-x / a^{2}},\) where \(A\) and \(a\) are constants. (a) Where is the particle most likely to be found? (b) Where is the probability per unit length half its maximum value?

Short Answer

Expert verified
The particle is most likely to be found at x = 0. The probability per unit length is half its maximum when x = \(0.5 * a^2 * ln(2)\).

Step by step solution

01

Evaluate the absolute square of the wave function

The absolute square of the wave function, |Ψ|², gives the probability density function. Applied to this problem, |Ψ|² = |\(A e^{-x/a^{2}}\)|² = \(A^{2} e^{-2x / a^{2}}\)
02

Analyze the maximum probability density

The maximum value of the function is the y-value when x = 0, because exponential decay functions achieve their maximum at x = 0. When x = 0, \(|Ψ|^2\) = \(A^2\). So the particle is most likely to be found at x = 0.
03

Solve for the x-value when the probability density is half its maximum

Setting |Ψ|² equal to 1/2 of its maximum value and solving for x gives the x-value when the probability per unit length is half its max: \(0.5 * A^2 = A^2 * e^{-2x / a^{2}}\) This reduces to 0.5 = e^{-2x / a^{2}}. Take the natural log of both sides to get ln(0.5) = -2x/ \(a^2\) . Multiply both sides by - \(a^2 / 2\) to solve for x, giving x = \(0.5 * a^2 * ln(2)\).
04

Generalize solution

The maximum probability density is at x = 0. The probability per unit length is half its maximum when x = \(0.5 * a^2 * ln(2)\). The exact numerical locations depend on the values of the constants A and a, which aren't given.

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