Question

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 $\mathit{L}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{25}\mathit{\mu }\mathit{m}$.

Short answer

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$66.3\times {10}^{-6}\mathrm{eV}$

Step-by-step solution

Total energy

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.

Identification of data

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

Finding the total energy

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

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

Finding the energies of the lowest five distinct states for a conduction electron moving in a cubical crystal

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$66.3\times {10}^{-6}\mathrm{eV}$