An electron (mass m) is contained in a rectangular corral of widths ${\mathbf{L}}_{\mathbf{x}}\mathbf{=}\mathbf{L}$and ${\mathbf{L}}_{\mathbf{Z}}\mathbf{=}\mathbf{2}\mathbf{L}$. (a) How many different frequencies of light could the electron emit or absorb if it makes a transition between a pair of the lowest five energy levels? What multiple of $\mathbf{h}\mathbf{/}\mathbf{8}{\mathbf{mL}}^{\mathbf{2}}$gives the (b) lowest, (c) second lowest, (d) third lowest, (e) highest, (f) second highest, and (g) third highest frequency?

The electron is a subatomic particle whose electric charge is negative one elementary charge. Electrons belong to the first generation of the lepton particle family and are generally thought to be elementary particles because they have no known components or substructure.

The quantized energies for an electron trapped in a two dimensional infinite potential well that forms a rectangular corral are

$E_{n_x,n_y}=\frac{h^2}{8m}\left(\frac{n_x^2}{L_x^2}+\frac{n_y^2}{L_y^2}\right)$

Here,

$n_x=$Quantum number for which the electron’s matter wave fits in well width $L_x$.

$n_y=$Quantum number for which the electron’s matter wave fits in well width $L_y$.

The energy level for an electron in the two-dimensional box,

$E_{n_x,n_y}=\frac{h^2}{8m}\left(\frac{n_x^2}{L_x^2}+\frac{n_y^2}{L_y^2}\right)$

Here,$L_y=L$and $L_z=2L$

$E_{n_x,n_y}=\frac{h^2}{8m}\left(\frac{n_x^2}{L_x^2}+\frac{n_y^2}{L_y^2}\right)$

So,

$$\begin{gathered} =\frac{h^2}{8m}\left(\frac{n_x^2}{L_x^2}+\frac{n_y^2}{{\left(2L\right)}^2}\right) \\ =\frac{h^2}{8m}\left(\frac{n_x^2}{L^2}+\frac{n_y^2}{4L^2}\right) \\ =\frac{h^2}{8m}\left(n_x^2+\frac{n_y^2}{4}\right) \end{gathered}$$

First five lowest possible energies:

$E_{n_x,n_y}==\frac{h^2}{8m{L}^2}\left(n_x^2+\frac{n_y^2}{4}\right)$

CaseI: (1,1)

$$\begin{gathered} E_{1,1}==\frac{h^2}{8m{L}^2}\left(1^2+\frac{1^2}{4}\right) \\ =1.25\left(\frac{h^2}{8m{L}^2}\right) \end{gathered}$$

Case IV: (2,1)

$$\begin{gathered} E_{1,1}==\frac{h^2}{8m{L}^2}\left(1^2+\frac{1^2}{4}\right) \\ =1.25\left(\frac{h^2}{8m{L}^2}\right)\left(2^2+\frac{1^2}{4}\right) \\ =4.25\left(\frac{h^2}{8m{L}^2}\right) \end{gathered}$$

Case V: (2,1)

$$\begin{gathered} E_{2,1}==\frac{h^2}{8m{L}^2}\left(2^2+\frac{1^2}{4}\right) \\ =4.25\left(\frac{h^2}{8m{L}^2}\right) \end{gathered}$$

Case VI: (2,2)

$$\begin{gathered} E_{2,2}==\frac{h^2}{8m{L}^2}\left(2^2+\frac{2^2}{4}\right) \\ =5\left(\frac{h^2}{8m{L}^2}\right) \end{gathered}$$

When energy difference is more, then the frequency is high, where as when energy difference is less, then the frequency is less

Different frequencies of light formed:

$$\begin{gathered} (E_{1,2}\to E_{1,1}):0.75 \\ (E_{1,3}\to E_{1,1}):2 \\ (E_{2,1}\to E_{1,1}):3 \\ (E_{2,2}\to E_{1,1}):3.75 \\ (E_{1,3}\to E_{1,2}):1.25 \\ (E_{2,1}\to E_{1,2}):2.25 \\ (E_{2,2}\to E_{1,2}):3 \\ (E_{2,1}\to E_{1,2}):1 \\ (E_{2,2}\to E_{1,3}):1.75 \\ (E_{2,2}\to E_{2,1}):0.75 \end{gathered}$$

There are $\boxed{eight}$possible frequencies.

Lowest possible frequency =$\boxed{0.75}$

Second lowest frequency=$\boxed{1}$

Third lowest frequency=$\boxed{1.25}$

Highest frequency=$\boxed{3.75}$

Second highest frequency=$\boxed{3}$

Third highest frequency= $\boxed{2.25}$.