Question

Obtain expression (5-23) from equation (5-22). Using $\mathbf{cos\theta }\mathbf{=}{\mathbf{cos}}^{\mathbf{2}}\left(\frac{1}{2}\theta \right)\mathbf{-}{\mathbf{sin}}^{\mathbf{2}}\left(\frac{1}{2}\theta \right)$and$\mathbf{sin\theta }\mathbf{=}\mathbf{2}\mathbf{sin}\left(\frac{1}{2}\theta \right)\mathbf{cos}\left(\frac{1}{2}\theta \right)$, first convert the argument of the cotangent from$\mathbf{kL}$to$\frac{\mathbf{1}}{\mathbf{2}}\mathbf{kL}$. Next, put the resulting equation in quadratic form, and then factor. Note that$\mathbf{\alpha }$is positive by definition.

Short answer

The obtained equation from the given reference is${\mathrm{U}}_0=\left\{\begin{array}{l}\frac{{\mathrm{\hslash }}^2{\mathrm{k}}^2}{2\mathrm{m}}{\sec }^2\frac{\mathrm{kL}}{2}\\ \frac{{\mathrm{\hslash }}^2{\mathrm{k}}^2}{2\mathrm{m}}{\csc }^2\frac{\mathrm{kL}}{2}\end{array}\right.$ .

Step-by-step solution

The concepts and formulas used to solve the given problem

Write the expression for value of $k$.

$\mathrm{k}=\frac{\sqrt{2\mathrm{mE}}}{\mathrm{\hslash }}$

Here, $k$ is expressed as constant, $m$ is mass, $E$is the energy and $\mathrm{\hslash }$is the reduced Planck's constant.

Rearrange the above expression for$E$.

$\begin{array}{l}{\mathrm{k}}^2=\frac{2\mathrm{mE}}{{\mathrm{\hslash }}^2}\\ \mathrm{E}=\frac{{\mathrm{\hslash }}^2{\mathrm{k}}^2}{2\mathrm{m}}\end{array}$

Write the expression for value of $\alpha$.

$\mathrm{\alpha }=\frac{\sqrt{2\mathrm{m}({\mathrm{U}}_0-\mathrm{E})}}{\mathrm{\hslash }}$

Here, $\mathrm{\alpha }$ is expressed as constant, $m$is mass, $U_0$is the potential, $E$is the energy and $\mathrm{\hslash }$is the reduced Planck's constant.

Write the expression for the energy quantization.

$2\mathrm{cotkL}=\frac{\mathrm{k}}{\mathrm{\alpha }}-\frac{\mathrm{\alpha }}{\mathrm{k}}$ …… (1)

Here,$\mathrm{L}$ is the depth of potential.

Substituting the different given values in the formulas stated above.

Substitute $\frac{\mathrm{coskL}}{\mathrm{sinkL}}$for $\mathrm{cotkL}$in equation (1).

$\frac{\mathrm{k}}{\mathrm{\alpha }}-\frac{\mathrm{\alpha }}{\mathrm{k}}=2\frac{\mathrm{coskL}}{\mathrm{sinkL}}$

$\frac{\mathrm{k}}{\mathrm{\alpha }}-\frac{\mathrm{\alpha }}{\mathrm{k}}=2\frac{{\cos }^2\left(\frac{1}{2}\mathrm{kL}\right)-{\sin }^2\left(\frac{1}{2}\mathrm{kL}\right)}{2\cos \left(\frac{1}{2}\mathrm{kL}\right)\sin \left(\frac{1}{2}\mathrm{kL}\right)}$

$\frac{\mathrm{k}}{\mathrm{\alpha }}-\frac{\mathrm{\alpha }}{\mathrm{k}}=\cot \left(\frac{1}{2}\mathrm{kL}\right)-\tan \left(\frac{1}{2}\mathrm{kL}\right)$

Multiply both sides by $\mathrm{\alpha k}$to above equation.

role="math" localid="1658312266794" $\begin{array}{l}{\mathrm{k}}^2-{\mathrm{\alpha }}^2=\mathrm{\alpha k}\left(\cot \left(\frac{1}{2}\mathrm{kL}\right)-\tan \left(\frac{1}{2}\mathrm{kL}\right)\right)\\ {\mathrm{\alpha }}^2+\mathrm{\alpha k}\left(\cot \left(\frac{1}{2}\mathrm{kL}\right)-\tan \left(\frac{1}{2}\mathrm{kL}\right)\right)-{\mathrm{k}}^2=0\end{array}$

Factorize the above equation.

$\left(\mathrm{\alpha }+\mathrm{kcot}\left(\frac{1}{2}\mathrm{kL}\right)\right)\left(\mathrm{\alpha }-\mathrm{kcot}\left(\frac{1}{2}\mathrm{kL}\right)\right)=0$

Solve the value of$\mathrm{\alpha }$from above equation

$\mathrm{\alpha }=-\mathrm{kcot}\left(\frac{1}{2}\mathrm{kL}\right),\mathrm{ktan}\left(\frac{1}{2}\mathrm{kL}\right)$

The value of $\mathrm{\alpha }$is positive. So

$\mathrm{\alpha }=\mathrm{ktan}\left(\frac{1}{2}\mathrm{kL}\right)$

Substitute$\frac{\sqrt{2\mathrm{m}({\mathrm{U}}_0-\mathrm{E})}}{\mathrm{\hslash }}$ for$\mathrm{\alpha }$and $\frac{\sqrt{2\mathrm{mE}}}{\mathrm{\hslash }}$for$\mathrm{k}$in above equation.

$\frac{\sqrt{2\mathrm{m}({\mathrm{U}}_0-\mathrm{E})}}{\mathrm{\hslash }}=\frac{\sqrt{2\mathrm{mE}}}{\mathrm{\hslash }}\tan \left(\frac{1}{2}\mathrm{kL}\right)$

Square on both sides of equation.

$\begin{array}{c}\frac{2\mathrm{m}({\mathrm{U}}_0-\mathrm{E})}{{\mathrm{\hslash }}^2}=\frac{2\mathrm{mE}}{{\mathrm{\hslash }}^2}{\tan }^2\left(\frac{1}{2}\mathrm{kL}\right)\\ {\mathrm{U}}_0-\mathrm{E}={\mathrm{Etan}}^2\left(\frac{1}{2}\mathrm{kL}\right)\\ {\mathrm{U}}_0=\mathrm{E}(1+{\tan }^2\left(\frac{1}{2}\mathrm{kL}\right)\\ {\mathrm{U}}_0={\mathrm{Esec}}^2\left(\frac{1}{2}\mathrm{kL}\right)\end{array}$

Substitute$\frac{{\mathrm{\hslash }}^2{\mathrm{k}}^2}{2\mathrm{m}}$ for$\mathrm{E}$in above equation.

${\mathrm{U}}_0=\frac{{\mathrm{\hslash }}^2{\mathrm{k}}^2}{2\mathrm{m}}{\sec }^2\left(\frac{1}{2}\mathrm{kL}\right)$

For negative value of $\cot \left(\frac{1}{2}\mathrm{kL}\right)$, the value of $\mathrm{\alpha }$will be positive.

$\mathrm{\alpha }=-\cot \left(\frac{1}{2}\mathrm{kL}\right)$

Substitute$\frac{\sqrt{2\mathrm{m}({\mathrm{U}}_0-\mathrm{E})}}{\mathrm{\hslash }}$ forrole="math" localid="1658312912484" $\alpha$and$\frac{\sqrt{2\mathrm{mE}}}{\mathrm{\hslash }}$for $k$in above equation.

$\frac{\sqrt{2\mathrm{m}({\mathrm{U}}_0-\mathrm{E})}}{\mathrm{\hslash }}=-\frac{\sqrt{2\mathrm{mE}}}{\mathrm{\hslash }}\cot \left(\frac{1}{2}\mathrm{kL}\right)$

Square on both sides of equation.

$\begin{array}{c}\frac{2\mathrm{m}({\mathrm{U}}_0-\mathrm{E})}{{\mathrm{\hslash }}^2}=\frac{2\mathrm{mE}}{{\mathrm{\hslash }}^2}{\cot }^2\left(\frac{1}{2}\mathrm{kL}\right)\\ {\mathrm{U}}_0-\mathrm{E}={\mathrm{Ecot}}^2\left(\frac{1}{2}\mathrm{kL}\right)\\ {\mathrm{U}}_0=\mathrm{E}\left(1+{\cot }^2\left(\frac{1}{2}\mathrm{kL}\right)\right)\\ {\mathrm{U}}_0={\mathrm{Ecsc}}^2\left(\frac{1}{2}\mathrm{kL}\right)\end{array}$

Substitute $\frac{{\mathrm{\hslash }}^2{\mathrm{k}}^2}{2\mathrm{m}}$for $\mathrm{E}$in above equation.

${\mathrm{U}}_0=\frac{{\mathrm{\hslash }}^2{\mathrm{k}}^2}{2\mathrm{m}}{\csc }^2\left(\frac{1}{2}\mathrm{kL}\right)$

Thus,${\mathrm{U}}_0=\left\{\begin{array}{l}\frac{{\mathrm{\hslash }}^2{\mathrm{k}}^2}{2\mathrm{m}}{\sec }^2\frac{\mathrm{kL}}{2}\\ \frac{{\mathrm{\hslash }}^2{\mathrm{k}}^2}{2\mathrm{m}}{\csc }^2\frac{\mathrm{kL}}{2}\end{array}\right.$ .