Question

A bound particle of massdescribed by the wave function

${\mathbf{\text{ψ(x)=Axe}}}^{{\mathbf{\text{-x}}}^{\mathbf{\text{2}}}{\mathbf{\text{/2b}}}^{\mathbf{\text{2}}}}$

What is the most probable location at which to find the particle?

Short answer

The most probably location to find the particle is at.$x=\pm b$

Step-by-step solution

Wave function at maximum.

The wave function is

$\psi \left(x\right)=Ax{e}^{-x^{2/2b^2}}$

The most probably location to find the particle is when the probability density reach maximum,

$\begin{array}{l}\rho =\psi \left(x\right){\psi }^{.}\left(x\right)\\ ={\left(A^2{x}^2{e}^{-x^{2/b^2}}\right)}^2\\ =A^2{x}^2{e}^{-x^{2/b^2}}\end{array}$

Derivative of the density.

The derivative of the density

$\begin{array}{c}\frac{d\rho }{dx}=A^2(2x{e}^{-x^{2/b^2}}+x^2{e}^{-x^{2/b^2}}(-\frac{2x}{b^2}))\\ =A^{2}2x{e}^{-x^{2/b^2}}(1-\frac{x^2}{b^2})\end{array}$

When it is zero, the probability is maximum or minimum,

$A^{2}2x{e}^{-x^{2/b^2}}(1-\frac{x^2}{b^2})=0$The solution is

$x=\pm b$or$0$.

At$x=0$, the density is$0$ , so it is a minimum.

Therefore, The most probably location to find the particle is at $x=\pm b$.