The deeper the finite well, the more state it holds. In fact, a new state, the, is added when the well’s depth${\mathit{U}}_{\mathbf{0}}$reaches${\mathit{h}}^{\mathbf{2}}{\mathbf{(}\mathbf{n}\mathbf{-}\mathbf{1}\mathbf{)}}^{\mathbf{2}}\mathbf{/}\mathbf{8}\mathit{m}{\mathit{L}}^{\mathbf{2}}$. (a) Argue that this should be the case based only on$\mathit{k}\mathbf{=}\sqrt{\mathbf{2}\mathbf{m}\mathbf{E}\mathbf{/}{\mathbf{h}}^{\mathbf{2}}}$, the shape of the wave inside, and the degree of penetration of the classically forbidden region expected for a state whose energy E is only negligibly below${\mathit{U}}_{\mathbf{0}}$. (b) How many states would be found up to this same “height” in an infinite well.

The wave function must be almost zero at the point it intersects with the wall because it has a very low exponential drop-off.

At$n=2$, the state can only fit one-half of a wavelength between the walls.

At $n=3$, the state can fit two half-wavelengths between the walls.

At $n=4$, the state can fit one complete wavelength and one half-wavelength.

Generally, each value of n can produce

Using this concept, we get:

$L=\frac{(n-1)}{2}\lambda$ Where$\lambda =\frac{2\pi }{k}$

$L=\frac{(n-1)}{2}\left(\frac{2\pi }{k}\right)$

$L=\frac{(n-1)\pi }{k}$

We need to express $k$using the kinetic energy,

${\rm E}=\frac{{\left(hk\right)}^2}{2m}$

$k=\sqrt{\frac{2m{\rm E}}{h^2}}$

Now, W relate $L$to$E$

$L=\frac{(n-1)\pi }{\frac{\sqrt{2m{\rm E}}}{h}}$

$L=\frac{(n-1)\pi h}{\sqrt{2mE}}$

$\sqrt{2mE}=\frac{(n-1)\pi h}{L}$

$E=\frac{{(n-1)}^2{\pi }^2{\left(\frac{h}{2\pi }\right)}^2}{2m{L}^2}$

$E=\frac{{(n-1)}^2{h}^2}{8m{L}^2}$

Hence, the minimum value of the potential energy, $U_0$, is equal to $\frac{{(n-1)}^2{h}^2}{8m{L}^2}$.

b). We can expect$(n-1)$ states below this energy, $\frac{{(n-1)}^2{h}^2}{8m{L}^2}$, given that the energy of the$(n-1)$ state in an infinite square well is equal to the energy $\frac{{(n-1)}^2{\pi }^2{h}^2}{2m{L}^2}$.