Question

Determine the particle’s most probable position.

Short answer

The most probable position for the particle with the given wave function is $\frac{3}{2a}$.

Step-by-step solution

Define the most probable position

A particle of mass m is represented by a wave function,

$\mathit{\psi }\left(x\right)\mathbf{=}\{\begin{array}{ll}2\sqrt{a^3}x{e}^{-ax}& x>0\\ 0& x<0\end{array}$

The most probable position of this particle is mathematically defined as

$x\text{'}=\underset{-\infty }{\overset{\infty }{\int }}{\psi }^2\left(x\right)xdx$

Determine the most probable position for the given wave function.

Expanding the integral into two parts,

$\begin{array}{rcl}x\text{'}& =& {\int }_{-\infty }^0{\psi }^2\left(x\right)xdx+{\int }_0^{\infty }{\psi }^2\left(x\right)xdx\\ & =& 0+{\int }_0^{\infty }{\left[2\sqrt{a^3}x{e}^{-ax}\right]}^{2}xdx\\ & =& {\int }_0^{\infty }\left[4a^3{x}^2{e}^{-2ax}\right]xdx\\ & =& 4a^3{\int }_0^{\infty }\left[x^3{e}^{-2ax}\right]dx\\ & & \end{array}$

Taking the integration part separately and solving it using By Parts method

$\begin{array}{rcl}I& =& {\int }_0^{\infty }\left[x^3{e}^{-2ax}\right]dx\\ & =& {\left.\frac{x^3{e}^{-2ax}}{-2a}-\frac{3x^2{e}^{-2ax}}{4a^2}+\frac{3x{e}^{-2ax}}{4a^3}-\frac{3e^{-2ax}}{8a^4}\right|}_0^{\infty }\\ & =& 0-\left(-\frac{3e^0}{8a^4}\right)\\ & =& \frac{3}{8a^4}\end{array}$

Inserting this result in the most probable expression,

$\begin{array}{rcl}x\text{'}& =& 4a^3\left(\frac{3}{8a^4}\right)\\ & =& \frac{3}{2a}\\ & & \end{array}$

Hence the most probable position for the particle with a given wave function is $\frac{3}{2a}$.