A rectangular corral of widths ${\mathbf{L}}_{\mathbf{x}}\mathbf{=}\mathbf{L}\mathbf{}\mathbf{and}\mathbf{}{\mathbf{L}}_{\mathbf{y}}\mathbf{=}\mathbf{2}\mathbf{L}$holds an electron. What multiple of ${\mathbf{h}}^{\mathbf{2}}\mathbf{/}\mathbf{8}{\mathbf{mL}}^{\mathbf{2}}$, where m is the electron mass, gives (a) the energy of the electron’s ground state, (b) the energy of its first excited state, (c) the energy of its lowest degenerate states, and (d) the difference between the energies of its second and third excited states?

The ground state of an electron, the energy level it normally occupies, is the state of lowest energy for that electron. There is also a maximum energy that each electron can have and still be part of its atom.

Two dimensional electron traps:

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 electrons matter wave fits in well width$L_x$.

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

(a)

Energy of electron,

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

Here, $L_x=L\mathrm{and}{L}_y=2L$and

So,

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

For the ground state,$n_x=1,n_y=1$

So,

role="math" localid="1661773510699" $$\begin{gathered} \frac{E_{n_x,n_y}}{{\displaystyle \frac{h^2}{8m{L}^2}}}=1+\frac{1}{4} \\ =1.25 \end{gathered}$$

Hence, the energy of the electron’s ground is 1.25.

(b)

For first excited state, $n_x=1\mathrm{and}{n}_y=2$

Hence,

role="math" localid="1661773643995" $$\begin{gathered} \frac{E_{n_x,n_y}}{{\displaystyle \frac{h^2}{8m{L}^2}}}={\left(1\right)}^2+\frac{1}{4}{\left(2\right)}^2 \\ =2 \\ =1.25 \end{gathered}$$

The energy of its first excited state is 1.25.

(c)

Lowest set states that are degenerates:

$$\begin{gathered} \left(n_x,n_y\right)=\left(1,4\right)\mathrm{and}\left(n_x,n_y\right)=\left(2,2\right) \\ \mathrm{For}1,4: \end{gathered}$$

$$\begin{gathered} \frac{E_{n_x,n_y}}{{\displaystyle \frac{h^2}{8m{L}^2}}}={\left(1\right)}^2+\frac{1}{4}{\left(4\right)}^2 \\ =5 \end{gathered}$$

For 2,2:

$$\begin{gathered} \frac{E_{n_x,n_y}}{{\displaystyle \frac{h^2}{8m{L}^2}}}={\left(2\right)}^2+\frac{1}{4}{\left(2\right)}^2 \\ =5 \end{gathered}$$

Hence, the energy of its lowest degenerate states is 5.

(d)

For second excited state:$n_x=1,n_y=3$

$$\begin{gathered} Hence, \\ \frac{E_{n_x,n_y}}{{\displaystyle \frac{h^2}{8m{L}^2}}}={\left(1\right)}^2+\frac{1}{4}{\left(3\right)}^2 \\ =3.25 \end{gathered}$$

For the excited state $n_x=2,n_y=1$

Hence,

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

Difference of the states =4.25-3.25=1

The difference between the energies of its second and third excited states is 1.