Question

An electron is trapped in a one-dimensional infinite potential well. Show that the energy difference$\mathbf{∆}\mathit{E}$ between its quantum levels n and n+2 is $\left(h^2/2m{L}^2\right)\left(n+1\right)$.

Short answer

It is proved that$∆E=\left(\frac{h^2}{2m{L}^2}\right)\left(n+1\right)$ .

Step-by-step solution

Describe the expression for Energy for the one-dimensional infinite potential well is given by

The Energy for the one-dimensional infinite potential well is given by,

${\mathbf{E}}_{\mathbf{n}}\mathbf{=}\left(\frac{h^2}{8{\mathrm{mL}}^2}\right){\mathbf{n}}^{\mathbf{2}}$

Show that the energy difference ∆E is (h2/2mL2)(n+1) 

Find the energy difference as follows.

$$\begin{gathered} ∆E=E_{n+2}-E_n \\ =\left(\frac{h^2}{8m{L}^2}\right){\left(n+2\right)}^2-\left(\frac{h^2}{8m{L}^2}\right)n^2 \\ =\left(\frac{h^2}{8m{L}^2}\right)\left[{\left(n+2\right)}^2-n^2\right] \\ =\left(\frac{h^2}{8m{L}^2}\right)\left(n^2+4n+4-n^2\right) \end{gathered}$$

Simplify further.

$$\begin{gathered} ∆E=\left(\frac{h^2}{8m{L}^2}\right)\left(n^2+4n+4-n^2\right) \\ =\left(\frac{h^2}{8m{L}^2}\right)\left(4\right)\left(n+1\right) \\ =\left(\frac{h^2}{8m{L}^2}\right)\left(n+1\right) \end{gathered}$$

Therefore, it is proved that$∆E=\left(\frac{h^2}{8m{L}^2}\right)\left(n+1\right)$ .