Question

Figure 5.15 shows that the allowed wave functions for a finite well whose depth ${\mathit{U}}_{\mathbf{0}}$was chosen to be$\mathbf{6}\mathit{\pi }{\mathit{\hslash }}^{\mathbf{2}}\mathbf{/}\mathit{m}{\mathit{L}}^{\mathbf{2}}$.

(a) Insert this value in equation (5-23), then using a calculator or computer, solve for the allowed value of $\mathit{k}\mathit{L}$, of which there are four.

(b) Using$\mathit{k}\mathbf{=}\frac{\sqrt{\mathbf{2}\mathbf{m}\mathbf{E}}}{\mathbf{\hslash }}$find corresponding values of E. Do they appear to agree with figure 5.15?

(c) Show that the chosen${\mathit{U}}_{\mathbf{0}}$implies that $\mathit{\alpha }\mathbf{\simeq }\sqrt{\frac{\mathbf{12}{\mathbf{\pi }}^{\mathbf{2}}}{{\mathbf{L}}^{\mathbf{2}}}\mathbf{-}{\mathbf{k}}^{\mathbf{2}}}$.

(d) Defining$\mathit{L}$and$\mathit{C}$to be 1 for convenience, plug your $\mathit{K}\mathit{L}$and $\mathit{\alpha }$values into the wave function given in exercise 46, then plot the results. ( Note: Your first and third $\mathit{K}\mathit{L}$values should correspond to even function of z, thus using the form with$\mathit{C}\mathit{O}\mathit{S}\mathbf{}\mathit{K}\mathit{Z}$, while the second and forth correspond to odd functions. Do the plots also agree with Figure 5.15?

Short answer

a) The allowed values $KL$are $2.650$, $7.821$,$5.272$, and $10.159$.

b) The expression for energy is $E=\frac{k^2{\hslash }^2}{2m}$, and yes, it agrees with figure 5.15.

c) It is proved that $\alpha =\sqrt{\frac{12{\pi }^2}{L^2}-k^2}$.

d)

yes, they agree with figure 5.15.

Step-by-step solution

Given data

The allowed wave functions for a finite well whose depth can be expressed as,

${\mathrm{U}}_0=\frac{6{\mathrm{\pi }}^2{\mathrm{\hslash }}^2}{{\mathrm{mL}}^2}$ (1)

The concepts and formula used to solve the given problem

The equation of potential energy can be written as,

${\mathbf{U}}_{\mathbf{0}}\mathbf{=}\mathbf{\{}\begin{array}{l}\frac{{\mathbf{\hslash }}^{\mathbf{2}}{\mathbf{k}}^{\mathbf{2}}}{\mathbf{2}\mathbf{m}}\mathbf{s}\mathbf{e}{\mathbf{c}}^{\mathbf{2}}\frac{\mathbf{k}\mathbf{L}}{\mathbf{2}}\\ \frac{{\mathbf{\hslash }}^{\mathbf{2}}{\mathbf{k}}^{\mathbf{2}}}{\mathbf{2}\mathbf{m}}\mathbf{c}\mathbf{s}{\mathbf{c}}^{\mathbf{2}}\frac{\mathbf{k}\mathbf{L}}{\mathbf{2}}\end{array}$ (2)

Here, ${\mathbf{U}}_{\mathbf{0}}$is the potential energy,$\mathit{K}$is energy constant,$\mathit{\hslash }$is reduced Planck's constant,$m$is the mass, and$\mathit{L}$is the depth of the well.

 a) Allowed values of KL

Substitute the values in the equation (2), and we get,

Part 1.

$\begin{array}{c}\frac{6{\mathrm{\pi }}^2{\mathrm{\hslash }}^2}{{\mathrm{mL}}^2}=\frac{{\mathrm{\hslash }}^2{\mathrm{k}}^2}{2\mathrm{m}}{\sec }^2\frac{\mathrm{kL}}{2}\\ \frac{12{\mathrm{\pi }}^2}{{\mathrm{k}}^2{\mathrm{L}}^2}={\sec }^2\frac{\mathrm{kL}}{2}\\ \mathrm{kLsec}\frac{\mathrm{kL}}{2}=\sqrt{12}\mathrm{\pi }\end{array}$

When we use the calculator to solve the above equation, we find the values as,

$\mathrm{kL}=2.650,3.868,7.821$

But 3.868 lies within $(\mathrm{\pi },2\mathrm{\pi })$, which fails the condition.

Part 2.

$\begin{array}{c}\frac{6{\mathrm{\pi }}^2{\mathrm{\hslash }}^2}{{\mathrm{mL}}^2}=\frac{{\mathrm{\hslash }}^2{\mathrm{k}}^2}{2\mathrm{m}}{\csc }^2\frac{\mathrm{kL}}{2}\\ \frac{12{\mathrm{\pi }}^2}{{\mathrm{k}}^2{\mathrm{L}}^2}={\csc }^2\frac{\mathrm{kL}}{2}\\ \mathrm{kLcsc}\frac{\mathrm{kL}}{2}=\sqrt{12}\mathrm{\pi }\end{array}$

When we use the calculator to solve the above equation, we find the values as,

$\mathrm{kL}=\text{5.272,7.911,10.159}$

But 7.911 lies within $(2\pi ,3\pi )$, which fails the condition.

Therefore, the allowed values $KL$are 2.650, $7.821$, $5.272$, and $10.159$.

(b) The concepts and formula used to solve the given problem

The expression for$K$is given as,

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

Here, $k$is energy constant, $E$is the energy.

Rearrange the above expression for E, and we get,

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

To compare it to figure 5.15, we can modify it as,

$\begin{array}{c}\mathrm{E}=\frac{{\mathrm{k}}^2{\mathrm{\hslash }}^2{\mathrm{L}}^{2}6{\mathrm{\pi }}^2}{2\times 6{\mathrm{\pi }}^2{\mathrm{mL}}^2}\\ \mathrm{E}=\frac{6{\mathrm{\pi }}^2{\mathrm{\hslash }}^2}{{\mathrm{mL}}^2}\frac{{\left(\mathrm{kL}\right)}^2}{12{\mathrm{\pi }}^2}\end{array}$

Substitute the values in the above equation from equation 1, and we get,

$\mathrm{E}={\mathrm{U}}_0\frac{{\left(\mathrm{kL}\right)}^2}{12{\mathrm{\pi }}^2}$

This expression is the fraction of the total well depth.

Thus, the expression for energy is $\mathrm{E}=\frac{{\mathrm{k}}^2{\mathrm{\hslash }}^2}{2\mathrm{m}}$, and yes, it agrees with figure 5.15.

c ) Show that  α≃12π2L2−k2

The expression for the value of $\mathrm{\alpha }$can be written as,

$\mathrm{\alpha }=\sqrt{\frac{2\mathrm{m}({\mathrm{U}}_0-\mathrm{E})}{{\mathrm{\hslash }}^2}}$ …… (3)

Here,$\alpha$ is the energy constant.

Substitute the values in the above equation, and we get,

$\begin{array}{c}\mathrm{\alpha }=\sqrt{\frac{2\mathrm{m}\left(\frac{6{\mathrm{\pi }}^2{\mathrm{\hslash }}^2}{{\mathrm{mL}}^2}-\frac{{\mathrm{k}}^2{\mathrm{\hslash }}^2}{2\mathrm{m}}\right)}{{\mathrm{\hslash }}^2}}\\ \mathrm{\alpha }=\sqrt{\frac{12{\mathrm{\pi }}^2}{{\mathrm{L}}^2}-{\mathrm{k}}^2}\end{array}$

Therefore, It is proved that$\mathrm{\alpha }=\sqrt{\frac{12{\mathrm{\pi }}^2}{{\mathrm{L}}^2}-{\mathrm{k}}^2}$

d ) Plots

The plots are given below as,

Yes, they agree with figure 5.15.