Question

Using equation (23), find the energy of a particle confined to a finite well whose walls are half the height of the ground-state infinite well energy, . (A calculator or computer able to solve equations numerically may be used, but this happens to be a case where an exact answer can be deduced without too much trouble.)

Short answer

The energy of a particle confined to a finite well whose walls are half the height of the ground-state infinite well energy is $\frac{{\pi }^2{\hslash }^2}{8m{L}^2}$.

Step-by-step solution

Given data

There is a finite well whose walls are half the height of the ground-state infinite well energy, E₁.

Ground state energy of infinite potential well and height of the finite potential well

The ground state energy of a particle of mass in an infinite well of width L is

$E=\frac{{\pi }^2{\hslash }^2}{2m{L}^2}$ .....(I)

Here $\hslash$ is the reduced Planck's constant.

The height of the finite potential well is

$U_0=\left\{\begin{array}{ll}\frac{{\hslash }^2{k}^2}{2m}se{c}^2\frac{kL}{2}& \left(n-1\right)\pi <kL<n\pi n=1,3,5...\\ \frac{{\hslash }^2{k}^2}{2m}cs{c}^2\frac{kL}{2}& \left(n-1\right)\pi <kL<n\pi n=2,4,6...\end{array}\right.$ .....(II)

Determining the energy of the finite well

The height of the well is half the ground state energy of an infinite potential well. Thus for n = 1

$U_0=\frac{E_1}{2}$

Substitute from equations (I) and (II) to get

$\begin{array}{rcl}\frac{{\hslash }^2{k}^2}{2m}{\sec }^2\frac{kL}{2}& =& \frac{{\pi }^2{\hslash }^2}{4m{L}^2}\\ k^2{\sec }^2\frac{kL}{2}& =& \frac{{\pi }^2}{2L^2}\\ {\cos }^2\frac{kL}{2}& =& \frac{2k^2{L}^2}{{\pi }^2}\end{array}$

This has a solution for

$k=\frac{\pi }{2L}$

The corresponding energy is

$\begin{array}{rcl}E& =& \frac{{\hslash }^2{k}^2}{2m}\\ & =& \frac{{\hslash }^2}{2m}\times \frac{{\pi }^2}{4L^2}\\ & =& \frac{{\pi }^2{\hslash }^2}{8m{L}^2}\end{array}$

Thus the required energy is $\frac{{\pi }^2{\hslash }^2}{8m{L}^2}$.