Question

Advance an argument based on $\mathit{p}\mathbf{=}\frac{\mathbf{h}}{\mathbf{\lambda }}$that there is no bound state in a half-infinite well unless ${\mathit{U}}_{\mathbf{0}}$is at least$\frac{{\mathbf{h}}^{\mathbf{2}}}{\mathbf{32}\mathbf{m}{\mathbf{L}}^{\mathbf{2}}}$. (Hint: What is the maximum wavelength possible within the well?)

Short answer

a). The particle must have$E<U_0$which means$E<U_0$, and so $U_0>\frac{h^2}{32m{L}^2}$.

b). The maximum wavelength allowed is$\lambda =4L$.

Step-by-step solution

Given Information

$$\begin{gathered} U\left(x\right)=\infty x\le 0 \\ U\left(x\right)=00<x<L \\ U\left(x\right)=U_{0}X\ge L \end{gathered}$$

Concept

The slower the wave function rises in the well, the wavelength increases.

However, the wave function cannot rise too slow because it has to meet up with a decaying exponential at $x>L$, where the classically forbidden region exists. We recall that it is required that the derivatives, if the wave function are continuous everywhere.

Specifically, the derivatives of the wave function in the free region $0<x<L$must match with those at the forbidden region, $x<L$.

Approach

A rising wave function as it approaches $x=L$has a positive derivative, whereas the decaying exponential has a negative derivative. For these two to match, the wave function in the free region must be either falling or flat when it approaches $x=L$.

In order to obtain the longest wavelength, we make the wave function flat when it reaches $X=L$. The sketch corresponds to $\frac{1}{4}$of a wavelength in the region between $x=0$and $x=L$, hence $\lambda =4L$is the maximum wavelength allowed.

Calculation

Given these information, we can argue that the smallest momentum and the smallest energy a bound particle can have are

$p=\frac{h}{{\lambda }_{m}ax}=\frac{h}{4L}$

$E=\frac{p^2}{2m}=\frac{h^2}{32m{L}^2}$

Hence, in order to be bound, the particle must have $E<U_0$which means $U_0>E_{min}$, and so $U_0>\frac{h^2}{32m{L}^2}.$.