91影视

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

${\mathrm{U}}_0=\frac{6{\mathrm{蟺}}^2{\mathrm{鈩�}}^2}{{\mathrm{mL}}^2}$ (1)

The equation of potential energy can be written as,

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

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

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

Part 1.

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

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 $(\mathrm{蟺},2\mathrm{蟺})$, which fails the condition.

Part 2.

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

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 $(2蟺,3蟺)$, which fails the condition.

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

The expression for$K$is given as,

$\mathrm{k}=\frac{\sqrt{2\mathrm{mE}}}{\mathrm{鈩�}}$

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

Rearrange the above expression for E, and we get,

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

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

$\begin{array}{c}\mathrm{E}=\frac{{\mathrm{k}}^2{\mathrm{鈩�}}^2{\mathrm{L}}^{2}6{\mathrm{蟺}}^2}{2脳6{\mathrm{蟺}}^2{\mathrm{mL}}^2}\\ \mathrm{E}=\frac{6{\mathrm{蟺}}^2{\mathrm{鈩�}}^2}{{\mathrm{mL}}^2}\frac{{\left(\mathrm{kL}\right)}^2}{12{\mathrm{蟺}}^2}\end{array}$

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

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

This expression is the fraction of the total well depth.

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

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

$\mathrm{伪}=\sqrt{\frac{2\mathrm{m}({\mathrm{U}}_0鈭�\mathrm{E})}{{\mathrm{鈩�}}^2}}$ 鈥︹�� (3)

Here,$伪$ is the energy constant.

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

$\begin{array}{c}\mathrm{伪}=\sqrt{\frac{2\mathrm{m}\left(\frac{6{\mathrm{蟺}}^2{\mathrm{鈩�}}^2}{{\mathrm{mL}}^2}鈭�\frac{{\mathrm{k}}^2{\mathrm{鈩�}}^2}{2\mathrm{m}}\right)}{{\mathrm{鈩�}}^2}}\\ \mathrm{伪}=\sqrt{\frac{12{\mathrm{蟺}}^2}{{\mathrm{L}}^2}鈭�{\mathrm{k}}^2}\end{array}$

Therefore, It is proved that$\mathrm{伪}=\sqrt{\frac{12{\mathrm{蟺}}^2}{{\mathrm{L}}^2}鈭�{\mathrm{k}}^2}$

The plots are given below as,

Yes, they agree with figure 5.15.