Question

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?

Short answer

a) The energy of the electron’s ground is 1.25.

b) The energy of its first excited state is 1.25.

c) The energy of its lowest degenerate states is 5.

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

Step-by-step solution

Introduction

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.

Concept

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 $L_y$.

Find the energy of the electron’s ground state

(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.

Find the energy of its first excited state

(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.

Find the energy of its lowest degenerate states

(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.

Find the difference between the energies of its second and third excited states

(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.