Question

Whereas an infinite well has an infinite number of bound states, a finite well does not. By relating the well height${\mathbf{\text{U}}}_{\mathbf{0}}$ to the kinetic energy and the kinetic energy (through $\lambda$) to n and L. Show that the number of bound states is given roughly by$\sqrt{\mathbf{8}\mathbf{m}{\mathbf{l}}^{\mathbf{2}}{\mathbf{U}}_{\mathbf{0}}\mathbf{/}{\mathbf{h}}^{\mathbf{2}}}$ (Assume that the number is large.)

Short answer

Hence, the number of bound states is given by$n_{\max }\approx \sqrt{\frac{8U_{0}m{L}^2}{h^2}}$ is proved

Step-by-step solution

Assumption

We assume that the wave function and energies correspond to the infinite well.

If,$E<U_0$ then inside the well, the energy is totally kinetic and is given by,

$\frac{h^2{k}^2}{2m}=\frac{n^2{\pi }^2{h}^2}{2m{L}^2}$

Calculation

The highest bound state exists when$E=U_0$. i.e., the difference$E-U_0\to 0$. For this state,

$\begin{array}{c}\frac{n_{\max }^2{\pi }^2{h}^2}{2m{L}^2}\approx U_0\\ n_{\max }\approx \sqrt{\frac{8U_{0}m{L}^2}{h^2}}\end{array}$

Hence the number of bounded states $n_{\max }\approx \sqrt{\frac{8U_{0}m{L}^2}{h^2}}$is proved.