A cubical box of widths ${\mathbf{L}}_{\mathbf{x}}\mathbf{=}{\mathbf{L}}_{\mathbf{y}}\mathbf{=}{\mathbf{L}}_{\mathbf{z}}\mathbf{=}\mathbf{L}$contains an electron. What multiple of ,${\mathbf{h}}^{\mathbf{2}}\mathbf{/}\mathbf{8}{\mathbf{mL}}^{\mathbf{2}}$where, m is the electron mass, is (a) the energy of the electron’s ground state, (b) the energy of its second excited state, and (c) the difference between the energies of its second and third excited states? How many degenerate states have the energy of (d) the first excited state and (e) the fifth excited state?

The energy equation for an electron trapped in a three-dimensional cell is given by,

${\mathbf{E}}_{{\mathbf{n}}_{\mathbf{x}}\mathbf{,}{\mathbf{n}}_{\mathbf{y}}\mathbf{,}{\mathbf{n}}_{\mathbf{z}}}\mathbf{=}\frac{{\mathbf{h}}^{\mathbf{2}}}{\mathbf{8}\mathbf{m}}\left(\frac{n_x^2}{L_x^2}+\frac{n_y^2}{L_y^2}+\frac{n_z^2}{L_z^2}\right)$ ….. (1)

Here, ${\mathbf{n}}_{\mathbf{x}}$is quantum number for which the electron’s matter wave fits in well width ${\mathbf{L}}_{\mathbf{x}}$,${\mathbf{n}}_{\mathbf{y}}$is quantum number for which the electron’s matter wave fits in well width ${\mathbf{L}}_{\mathbf{y}}$, ${\mathbf{n}}_{\mathbf{z}}$is quantum number for which the electron’s matter wave fits in well width ${\mathbf{L}}_{\mathbf{z}}$, h is plank constant, and m is mass of the electron.

The first excited state occur at ${\mathrm{n}}_{\mathrm{x}}=1,{\mathrm{n}}_{\mathrm{y}}=1,\mathrm{and}{\mathrm{n}}_{\mathrm{z}}=1$.

Substitute L for ${\mathrm{L}}_{\mathrm{x}},\mathrm{L}\mathrm{for}{\mathrm{L}}_{\mathrm{y}},\mathrm{L}\mathrm{for}{\mathrm{L}}_{\mathrm{z}},1\mathrm{for}{\mathrm{n}}_{\mathrm{x}},1\mathrm{for}{\mathrm{n}}_{\mathrm{y}},$and 1 for $n_z$in equation (1).

$$\begin{gathered} {\mathrm{E}}_{1,1,1}=\frac{{\mathrm{h}}^2}{8\mathrm{m}}\left(\frac{1^2}{{\mathrm{L}}^2}+\frac{1^2}{{\mathrm{L}}^2}+\frac{1^2}{{\mathrm{L}}^2}\right) \\ =\frac{{\mathrm{h}}^2}{8{\mathrm{mL}}^2}\left(1+1+1\right) \\ =3\left(\frac{{\mathrm{h}}^2}{8{\mathrm{mL}}^2}\right) \end{gathered}$$

Therefore, the multiple of $\frac{{\mathrm{h}}^2}{8{\mathrm{mL}}^2}$is 3 .

The second excited state occur at ${\mathrm{n}}_{\mathrm{x}}=2,{\mathrm{n}}_{\mathrm{y}}=2,\mathrm{and}{\mathrm{n}}_{\mathrm{z}}=1$.

Substitute L for role="math" localid="1661931817535" ${\mathrm{L}}_{\mathrm{x}},\mathrm{L}\mathrm{for}{\mathrm{L}}_{\mathrm{y}},\mathrm{L}\mathrm{for}{\mathrm{L}}_{\mathrm{z}},2\mathrm{for}{\mathrm{n}}_{\mathrm{x}},2\mathrm{for}{\mathrm{n}}_{\mathrm{y}}$, and 1 for $n_z$in equation (1).

$$\begin{gathered} {\mathrm{E}}_{2,2,1}=\frac{{\mathrm{h}}^2}{8\mathrm{m}}\left(\frac{2^2}{{\mathrm{L}}^2}+\frac{2^2}{{\mathrm{L}}^2}+\frac{1^2}{{\mathrm{L}}^2}\right) \\ =\frac{{\mathrm{h}}^2}{8{\mathrm{mL}}^2}\left(\frac{4}{{\mathrm{L}}^2}+\frac{4}{{\mathrm{L}}^2}+\frac{1}{{\mathrm{L}}^2}\right) \\ =\frac{{\mathrm{h}}^2}{8{\mathrm{mL}}^2}\left(4+4+1\right) \\ =9\left(\frac{{\mathrm{h}}^2}{8{\mathrm{mL}}^2}\right) \end{gathered}$$

Therefore, the multiple of $\frac{{\mathrm{h}}^2}{8{\mathrm{mL}}^2}$is 9 .

The energy of the third exited state occurs at $\left(n_x,n_y\right)=\left(1,1\right)$and $n_z=3$.

Substitute L for ${\mathrm{L}}_{\mathrm{x}},\mathrm{L}\mathrm{for}{\mathrm{L}}_{\mathrm{y}},\mathrm{L}\mathrm{for}{\mathrm{L}}_{\mathrm{z}},1\mathrm{for}{\mathrm{n}}_{\mathrm{x}},1\mathrm{for}{\mathrm{n}}_{\mathrm{y}}$and 3 for ${\mathrm{n}}_{\mathrm{z}}$in equation (1).

$$\begin{gathered} {\mathrm{E}}_{1,1,3}=\frac{{\mathrm{h}}^2}{8\mathrm{m}}\left(\frac{1^2}{{\mathrm{L}}^2}+\frac{1^2}{{\mathrm{L}}^2}+\frac{3^2}{{\mathrm{L}}^2}\right) \\ =\frac{{\mathrm{h}}^2}{8{\mathrm{mL}}^2}\left(\frac{1}{{\mathrm{L}}^2}+\frac{1}{{\mathrm{L}}^2}+\frac{9}{{\mathrm{L}}^2}\right) \\ =\frac{{\mathrm{h}}^2}{8{\mathrm{mL}}^2}\left(1+1+9\right) \\ =11\left(\frac{{\mathrm{h}}^2}{8{\mathrm{mL}}^2}\right) \end{gathered}$$

The difference in energy between the ground state and the second exited state is calculated as follows.

$$\begin{gathered} E_{1,1,,3}-E_{2,2,1}=11\left(\frac{{\mathrm{h}}^2}{8{\mathrm{mL}}^2}\right)-9\left(\frac{{\mathrm{h}}^2}{8{\mathrm{mL}}^2}\right) \\ =2\left(\frac{{\mathrm{h}}^2}{8{\mathrm{mL}}^2}\right) \end{gathered}$$

Therefore, the difference in energy between the ground state and the second exited state is,

$E_{1,1,,3}-E_{2,2,1}=2\left(\frac{{\mathrm{h}}^2}{8{\mathrm{mL}}^2}\right)$

The first exited state occur at $(n_x,n_y,n_z)=(2,\hspace{0.33em}1,\hspace{0.33em}1)$, $(n_x,n_y,n_z)=(1,\hspace{0.33em}2,\hspace{0.33em}1)$, and $(n_x,n_y,n_z)=(1,\hspace{0.33em}1,\hspace{0.33em}2)$.

Therefore, the number of degenerate states are 3.

The first exited state occur at.

$(n_x,n_y,n_z)=(1,\hspace{0.33em}2,\hspace{0.33em}3),\hspace{0.33em}(1,\hspace{0.33em}3,\hspace{0.33em}2),\hspace{0.33em}(2,\hspace{0.33em}3,\hspace{0.33em}1),\hspace{0.33em}(2,\hspace{0.33em}1,\hspace{0.33em}3),\hspace{0.33em}(3,\hspace{0.33em}1,\hspace{0.33em}2),\hspace{0.33em}(3,\hspace{0.33em}2,\hspace{0.33em}1)$

Therefore, the number of degenerate states are 6.