Consider a conduction electron in a cubical crystal of a conducting material. Such an electron is free to move throughout the volume of the crystal but cannot escape to the outside. It is trapped in a three-dimensional infinite well. The electron can move in three dimensions so that its total energy is given by

$\mathit{E}\mathbf{=}\frac{{\mathbf{h}}^{\mathbf{2}}}{\mathbf{8}{\mathbf{L}}^{\mathbf{2}}\mathbf{m}}\left(n_1^2+n_2^2+n_3^2\right)$

in whichare positive integer values. Calculate the energies of the lowest five distinct states for a conduction electron moving in a cubical crystal of edge length $\mathit{L}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{25}\mathit{\mu }\mathit{m}$.

Total Energy is the sum of all final energy used at a certain branch or variable. Because energy data is entered directly, total energy can be distinguished from final energy intensity.

Here we have given that,

Total energy is given by$E=\frac{h^2}{8L^{2}m}\left(n_1^2+n_2^2+n_3^2\right)$

Edge length of cubical crystal is$L=0.25\mu m$

Mass of electron is$\mathrm{m}=9.11\times {10}^{-31}\mathrm{kg}$

Plank’s constant is$\mathrm{h}=6.626\times {10}^{-34}\mathrm{J}.\mathrm{s}$

Here we have given that total energy is given by,

$E=\frac{h^2}{8L^{2}m}\left(n_1^2+n_2^2+n_3^2\right)$

Now, by substituting the numerical values in above equation we get,

$$\begin{gathered} \mathrm{E}=\frac{{\left(6.626\times {10}^{-34}\mathrm{J}.\mathrm{s}\right)}^2}{8{\left(0.25\times {10}^{-6}\mathrm{m}\right)}^2\left(9.11\times {10}^{-31}\mathrm{kg}\right)}\left({\mathrm{n}}_1^2+{\mathrm{n}}_2^2+{\mathrm{n}}_3^2\right) \\ =9.63\times {10}^{-25}\mathrm{J}\left({\mathrm{n}}_1^2+{\mathrm{n}}_2^2+{\mathrm{n}}_3^2\right) \end{gathered}$$

Now, we know that$1\mathrm{eV}=1.6\times {10}^{=19}\mathrm{J}$

So, the value of E becomes

$$\begin{gathered} \mathrm{E}=\frac{9.63\times {10}^{-25}}{1.6\times {10}^{-18}\mathrm{J}/\mathrm{eV}}\left({\mathrm{n}}_1^2+{\mathrm{n}}_2^2+{\mathrm{n}}_3^2\right) \\ =\left(6.024\times {10}^{-6}\mathrm{eV}\right)\left({\mathrm{n}}_1^2+{\mathrm{n}}_2^2+{\mathrm{n}}_3^2\right) \end{gathered}$$

Now, the lowest 5 states are,

1. ${\mathrm{n}}_1={\mathrm{n}}_2={\mathrm{n}}_3=1$
   
2. ${\mathrm{n}}_1={\mathrm{n}}_2=1,{\mathrm{n}}_3=2$
   
3. ${\mathrm{n}}_1=1,{\mathrm{n}}_2={\mathrm{n}}_3=2$
   
4. ${\mathrm{n}}_1={\mathrm{n}}_2={\mathrm{n}}_3=2$
   
5. ${\mathrm{n}}_1={\mathrm{n}}_2=1,{\mathrm{n}}_3=3$
   

Now, the energy for$n_1=n_2=n_3=1$is given by,

$$\begin{gathered} {\mathrm{E}}_{1,1,1}=\left(6.024\times {10}^{-6}\mathrm{eV}\right)\left({\left(1\right)}^2+{\left(1\right)}^2+{\left(1\right)}^2\right) \\ =18.1\times {10}^{-6}\mathrm{eV} \end{gathered}$$

The energy for$n_1=n_2=1,n_3=2$is given by,

$$\begin{gathered} {\mathrm{E}}_{1,1,2}=\left(6.024\times {10}^{-6}\mathrm{eV}\right)\left({\left(1\right)}^2+{\left(1\right)}^2+{\left(2\right)}^2\right) \\ =36.2\times {10}^{-6}\mathrm{eV} \end{gathered}$$

The energy for$n_1=1,n_2=n_3=2$is given by,

$$\begin{gathered} {\mathrm{E}}_{1,2,2}=\left(6.024\times {10}^{-6}\mathrm{eV}\right)\left({\left(1\right)}^2+{\left(2\right)}^2+{\left(2\right)}^2\right) \\ =54.3\times {10}^{-6}\mathrm{eV} \end{gathered}$$

The energy for$n_1=n_2=n_3=2$is given by,

$$\begin{gathered} {\mathrm{E}}_{2,2,2}=\left(6.024\times {10}^{-6}\mathrm{eV}\right)\left({\left(2\right)}^2+{\left(2\right)}^2+{\left(2\right)}^2\right) \\ =72.4\times {10}^{-6}\mathrm{eV} \end{gathered}$$

The energy foris given by,

$$\begin{gathered} {\mathrm{E}}_{1,1,2}=\left(6.024\times {10}^{-6}\mathrm{eV}\right)\left({\left(1\right)}^2+{\left(1\right)}^2+{\left(3\right)}^2\right) \\ =66.3\times {10}^{-6}\mathrm{eV} \end{gathered}$$

Hence, the energies of the lowest five distinct states for a conduction electron moving in a cubical crystal are$18.1\mathrm{x}{10}^{-6}\mathrm{eV},36.2\times {10}^{-6}\mathrm{eV},54.3\times {10}^{-6}\mathrm{eV},72.4\times {10}^{-6}\mathrm{eV}$ and$66.3\times {10}^{-6}\mathrm{eV}$