An electron is in a three-dimensional box with side lengths ${\mathbf{L}}_{\mathbf{x}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{600}\mathbf{}\mathbf{nm}$and ${\mathbf{L}}_{\mathbf{y}}\mathbf{=}{\mathbf{L}}_{\mathbf{z}}\mathbf{=}\mathbf{2}{\mathbf{L}}_{\mathbf{x}}$. What are the quantum numbers ${\mathbf{n}}_{\mathbf{x}}\mathbf{,}{\mathbf{n}}_{\mathbf{y}}$, and ${\mathbf{n}}_{\mathbf{z}}$and the energies, in ev , for the four lowest energy levels? What is the degeneracy of each (including the degeneracy due to spin)?

In the three-dimensional box with the sides of length ${\mathbf{L}}_{\mathbf{x}}\mathbf{,}{\mathbf{L}}_{\mathbf{z}}\mathbf{}\mathbf{and}\mathbf{}{\mathbf{L}}_{\mathbf{z}}$the energy levels is:

localid="1668241819620" ${\mathbf{E}}_{{\mathbf{n}}_{\mathbf{x}}\mathbf{,}{\mathbf{n}}_{\mathbf{y}}\mathbf{,}{\mathbf{n}}_{\mathbf{z}}}\mathbf{=}\mathbf{(}\frac{{\mathbf{n}}_{\mathbf{x}}^{\mathbf{2}}}{{\mathbf{L}}_{\mathbf{x}}^{\mathbf{2}}}\mathbf{+}\frac{{\mathbf{n}}_{\mathbf{y}}^{\mathbf{2}}}{{\mathbf{L}}_{\mathbf{y}}^{\mathbf{2}}}\mathbf{+}\frac{{\mathbf{n}}_{\mathbf{z}}^{\mathbf{2}}}{{\mathbf{L}}_{\mathbf{z}}^{\mathbf{2}}}\mathbf{)}\frac{{\mathbf{\pi }}^{\mathbf{2}}{\sout{\mathbf{h}}}^{\mathbf{2}}}{\mathbf{2}\mathbf{m}}$……………(1)

Given that,

$$\begin{gathered} L_x=0.600nm \\ {L}_y=L_z \\ =2L_x \end{gathered}$$

The mass of a photon${\mathrm{m}}_{\mathrm{c}}=9.11\times {10}^{31}\mathrm{kg}$

(a)

The ground level is$\left(n_x{n}_y,n_x\right)=\left(1,1,1\right)$

The energy level of an electron in the three-dimensional box is:

$$\begin{gathered} {\mathrm{E}}_{1,1,1}=\left(4{\mathrm{n}}_{\mathrm{x}}^2+{\mathrm{n}}_{\mathrm{y}}^2+{\mathrm{n}}_{\mathrm{z}}^2\right)\frac{{\mathrm{\pi }}^2{\mathrm{h}}^2}{8{\mathrm{mL}}_{\mathrm{x}}^2} \\ =(4{\left(1\right)}^2+{\left(1\right)}^2+{\left(1\right)}^2)\left[\left\{\frac{\mathrm{\pi }2(1.055\times 10-34\mathrm{J}.\mathrm{s})2}{8\times \left(9.11\times 10-31\mathrm{kg}\right)\times \left(0.600\times 10-9\mathrm{m}\right)2}\right\}\left(\frac{1\mathrm{eV}}{1.60\times 10-19\mathrm{J}}\right)\right] \\ =(4{\left(1\right)}^2+{\left(1\right)}^2+{\left(1\right)}^2)\left[0.2619\mathrm{eV}\right] \\ =1.569\mathrm{eV} \end{gathered}$$

Hence, the energy in ground level is 1.569 eV and the degeneracy of ground level is 2 states.

The energy level of an electron in the three-dimensional box is:

(b)

The second level is$\left(n_x,n_y,n_z\right)=\left(1,2,1\right)\left(1,1,2\right)$

$$\begin{gathered} {\mathrm{E}}_{1,2,1}={\mathrm{E}}_{1,1,2} \\ =\left(4{\left(1\right)}^2+{\left(2\right)}^2+{\left(2\right)}^2\right)\left[\left\{\frac{{\mathrm{\pi }}^2{\left(1.055\times {10}^{-34}\mathrm{J}.\mathrm{s}\right)}^2}{8\times \left(9.11\times {10}^{-31}\mathrm{kg}\right)\times \left(0.60\times {10}^{-19}\mathrm{J}\right)}\right\}\left\{\frac{1\mathrm{eV}}{1.60\times {10}^{-19}\mathrm{J}}\right\}\right] \\ =2.353\mathrm{eV} \end{gathered}$$

The degeneracy of ground level is $2\times 2=4$states

(c)

The third level is$\left(n_x,n_y,n_z\right)=\left(1,2,2\right)$

$$\begin{gathered} {\mathrm{E}}_{1,2,1}=\left(4{\left(1\right)}^2+{\left(2\right)}^2+{\left(2\right)}^2\right)\left[\left\{\frac{{\mathrm{\pi }}^2{\left(1.055\times {10}^{-34}\mathrm{J}.\mathrm{s}\right)}^2}{8\times \left(9.11\times {10}^{-31}\mathrm{kg}\right)\times \left(0.60\times {10}^{-19}\mathrm{J}\right)}\right\}\left\{\frac{1\mathrm{eV}}{1.60\times {10}^{-19}\mathrm{J}}\right\}\right] \\ =3.138\mathrm{eV} \end{gathered}$$

The degeneracy of ground level is 2 states.

Hence, the energy in the second level is 2.353 eV and the degeneracy of the second level is 4 states. Also, the energy in the third level is 3.138 eV and the degeneracy of the third level is 2 states.

(d)

The energy level of an electron in the three-dimensional box is:

The second level is$\left({\mathrm{n}}_{\mathrm{x}},{\mathrm{n}}_{\mathrm{y}},{\mathrm{n}}_{\mathrm{z}}\right)=\left(1,3,1\right)\left(1,1,3\right)$

$$\begin{gathered} {\mathrm{E}}_{1,3,1}={\mathrm{E}}_{1,1,3} \\ =\left(4{\left(1\right)}^2+{\left(3\right)}^2+{\left(1\right)}^2\right)\left[\left\{\frac{{\mathrm{\pi }}^2{\left(1.055\times {10}^{-34}\mathrm{J}.\mathrm{s}\right)}^2}{8\times \left(9.11\times {10}^{-31}\mathrm{kg}\right)\times \left(0.600\times {10}^{-19}\mathrm{J}\right)}\right\}\left\{\frac{1\mathrm{eV}}{1.60\times {10}^{-19}\mathrm{J}}\right\}\right] \\ =3.661\mathrm{eV} \end{gathered}$$

Hence, the energy in the fourth level is 3.661 eV and the degeneracy of the fourth level is 4 states.