To show that the potential energy of finite well is $\mathbf{U}\mathbf{=}\frac{{\mathbf{h}}^{\mathbf{2}}{\mathbf{(}\mathbf{n}\mathbf{-}\mathbf{1}\mathbf{)}}^{\mathbf{2}}}{\mathbf{8}{\mathbf{mL}}^{\mathbf{2}}}$

The potential energy is the stored energy that depends upon the relative position of various parts of a system.

The given potential in the finite well is:

$\mathrm{U}=\left\{\begin{array}{l}\begin{array}{ll}\frac{{\mathrm{\hslash }}^2{\mathrm{k}}^2}{2\mathrm{m}}{\sec }^2\frac{\mathrm{kL}}{2}& (\mathrm{n}-1)\mathrm{\pi }<\mathrm{kL}<\mathrm{n\pi }\mathrm{n}=1,3,5,\dots \end{array}\\ \begin{array}{ll}\frac{{\mathrm{\hslash }}^2{\mathrm{k}}^2}{2\mathrm{m}}{\csc }^2\frac{\mathrm{kL}}{2}& (\mathrm{n}-1)\mathrm{\pi }<\mathrm{kL}<\mathrm{n\pi }\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{n}=2,4,6....\end{array}\end{array}\right.$

The formula for the potential for even $\mathrm{n}$is given by:

$\mathrm{U}=\frac{{\mathrm{\hslash }}^2{\mathrm{k}}^2}{2\mathrm{m}}{\csc }^2\frac{\mathrm{kL}}{2}(\mathrm{n}-1)\mathrm{\pi }<\mathrm{kL}<\mathrm{n\pi }\dots \mathrm{...}\text{(1)}$

Here,$U$ is the potential, $\hslash$is the reduced Planck's constant, $k$is expressed as constant,$m$ is the mass and $L$is the depth of the potential well.

The expression for a minimum value of $k$.

$\mathrm{k}=\frac{(\mathrm{n}-1)\mathrm{\pi }}{\mathrm{L}}$

Here, $n$is the number of states.

Calculation:

Substitute$\frac{(\mathrm{n}-1)\mathrm{\pi }}{\mathrm{l}}$for$k$in equation (1).

$\mathrm{U}=\frac{{\mathrm{\hslash }}^2}{2\mathrm{m}}\left(\frac{{(\mathrm{n}-1)}^2{\mathrm{\pi }}^2}{{\mathrm{L}}^2}\right){\csc }^2\frac{\mathrm{L}}{2}\left(\frac{(\mathrm{n}-1)\mathrm{\pi }}{\mathrm{L}}\right)$

$\mathrm{U}==\frac{{\mathrm{\hslash }}^2{\mathrm{\pi }}^2{(\mathrm{n}-1)}^2}{2{\mathrm{mL}}^2}{\csc }^2\left(\frac{(\mathrm{n}-1)\mathrm{\pi }}{2}\right)$

Substitute$1$ for the minimum value of${\csc }^2\left(\frac{(n-1)\pi }{2}\right)$ in above equation.

$\mathrm{U}=\frac{{\mathrm{\hslash }}^2{\mathrm{\pi }}^2{(\mathrm{n}-1)}^2}{2{\mathrm{mL}}^2}$

Substitute $\frac{\mathrm{h}}{2\mathrm{\pi }}$for$\hslash$ in the above equation.

$\mathrm{U}=\frac{{\mathrm{h}}^2{\mathrm{\pi }}^2{(\mathrm{n}-1)}^2}{8{\mathrm{mL}}^2}$

Thus, the potential energy of infinite well is$\frac{{\mathrm{h}}^2{\mathrm{\pi }}^2{(\mathrm{n}-1)}^2}{8{\mathrm{mL}}^2}$ .