To determine the two bound state energies for the well.

Given Information:

${\mathrm{U}}_0=4\left(\frac{{\mathrm{\pi }}^2{\mathrm{\hslash }}^2}{2{\mathrm{mL}}^2}\right)$

Formula Used:

The result of the exercise 40 is

$\sqrt{\mathrm{E}}\cot \left(\frac{\sqrt{2\mathrm{mE}}}{\mathrm{h}}\mathrm{L}\right)=-\sqrt{{\mathrm{U}}_0-\mathrm{E}}$

Here, $\mathrm{E}$ is the energy,$m$ is the mass, $h$ is the reduced Planck's constant, $L$ is the width of the infinite well and $U_0$ is the potential energy of the particle.

Multiply the above equation by $\frac{\sqrt{2\mathrm{mE}}}{\mathrm{h}}\mathrm{L}$ on both sides

This modifies the equation to:

$\begin{array}{c}\frac{\sqrt{2\mathrm{m}}}{\mathrm{h}}\mathrm{L}\cdot \sqrt{\mathrm{E}}\cot \left(\frac{\sqrt{2\mathrm{mE}}}{\mathrm{h}}\mathrm{L}\right)=\frac{\sqrt{2\mathrm{m}}}{\mathrm{h}}\mathrm{L}\cdot (-\sqrt{{\mathrm{U}}_0-\mathrm{E}})\\ \frac{\sqrt{2\mathrm{mE}}}{\mathrm{h}}\mathrm{Lcot}\left(\frac{\sqrt{2\mathrm{mE}}}{\mathrm{h}}\mathrm{L}\right)=-\sqrt{\frac{2{\mathrm{mU}}_0{\mathrm{L}}^2-2{\mathrm{mEL}}^2}{{\mathrm{h}}^2}}\end{array}$

Make the substitution,$\mathrm{x}=\frac{\sqrt{2\mathrm{mE}}}{\mathrm{h}}\mathrm{L}$

$\mathrm{xcot}\left(\mathrm{x}\right)=-\sqrt{\frac{2{\mathrm{mU}}_0{\mathrm{L}}^2}{{\mathrm{\hslash }}^2}-{\mathrm{x}}^2}$ …… (1)

The potential energy${\mathrm{U}}_0$ is

${\mathrm{U}}_0=4\left(\frac{{\mathrm{\pi }}^2{\mathrm{\hslash }}^2}{2{\mathrm{mL}}^2}\right)$

Substitute the above equation in the equation (1) and simplify.

$\begin{array}{l}\mathrm{xcot}\left(\mathrm{x}\right)=-\sqrt{\frac{2\mathrm{m}\left(4\left(\frac{{\mathrm{\pi }}^2{\mathrm{h}}^2}{2{\mathrm{mL}}^2}\right)\right){\mathrm{L}}^2}{{\mathrm{h}}^2}-{\mathrm{x}}^2}\\ \mathrm{xcot}\left(\mathrm{x}\right)=-\sqrt{4{\mathrm{\pi }}^2-{\mathrm{x}}^2}\end{array}$

This is a very nice function to plug into computer program of choice.

Solutions are:

$\begin{array}{c}\mathrm{x}=2.698\text{and}5.284\\ \text{For}\mathrm{x}=2.698\\ 2.698=\frac{\sqrt{2{\mathrm{mE}}_1}}{\mathrm{h}}\mathrm{L}\\ {\mathrm{E}}_1{\left(2.698\right)}^2={\left(\frac{\mathrm{hh}}{\sqrt{2\mathrm{mL}}}\right)}^2\\ =7.28\frac{{\mathrm{h}}^2}{2{\mathrm{mL}}^2}\end{array}$

And for$\mathrm{x}=5.284$we get

$\begin{array}{c}5.284=\frac{\sqrt{2{\mathrm{mE}}_2}}{\mathrm{h}}\mathrm{L}\\ {\mathrm{E}}_2{\left(5.284\right)}^2={\left(\frac{\mathrm{h}}{\sqrt{2\mathrm{mL}}}\right)}^2\\ =27.9\frac{{\mathrm{h}}^2}{2{\mathrm{mL}}^2}\end{array}$

Hence, the two bound state energies for a well is ${\mathrm{E}}_1=7.28\frac{{\mathrm{h}}^2}{2{\mathrm{mL}}^2}$ and ${\mathrm{E}}_2=27.9\frac{{\mathrm{h}}^2}{2{\mathrm{mL}}^2}$.