An electron confinedtoa cubic 3D infinite well 1 nu on aside.

1. What are thethree lowest differentenergies?
2. To how many different states do these three energies correspond?

The three quantum numbers describe the states of the cubic square well are $\left(n_x,n_y,n_z\right)$and the energy for a given set of quantum numbers.

Then write the relation as below.

${\mathit{E}}_{{\mathbf{n}}_{\mathbf{x}}\mathbf{,}{\mathbf{n}}_{\mathbf{y}}\mathbf{,}{\mathbf{n}}_{\mathbf{z}}}\mathbf{=}\left(n_x^2+n_y^2+n_z^2\right)\frac{{\mathbf{\pi }}^{\mathbf{2}}{\mathbf{h}}^{\mathbf{2}}}{\mathbf{2}\mathbf{m}{\mathbf{L}}^{\mathbf{2}}}$

For the ground state $\left(n_x{n}_y,n_z\right)=\left(1,1,1\right)$which has the lowest energy.

The corresponding energy of the ground state is,

$E_{1,1,1}=\left(1^2+1^2+1^2\right)\frac{{\mathrm{\pi }}^2{\mathrm{h}}^2}{2m{L}^2}$

Substitute $1.054\times {10}^{-34}\mathrm{J}.\mathrm{s}$ for h , $9.1094\times {10}^{-31}$kg for m, and 1nm for L in the above equation.

$\begin{array}{r}{E}_{1,1,1}=\left(1^2+1^2+1^2\right)\frac{{\pi }^2{h}^2}{2m{L}^2}\\ =3\frac{(3.14159{)}^2{\left(1.054\times {10}^{-34}\mathrm{J}.\mathrm{s}\right)}^2}{2\left(9.1094\times {10}^{-31}\mathrm{kg}\right)(1\mathrm{nm}{)}^2}\\ =3\frac{(3.14159{)}^2{\left(1.054\times {10}^{-34}\mathrm{J}.\mathrm{s}\right)}^2}{2\left(9.1094\times {10}^{-31}\mathrm{kg}\right)\left(1\mathrm{nm}\right){\left(\frac{{10}^{-9}\mathrm{m}}{\mathrm{nm}}\right)}^2}\\ =3\left(6.0247\times {10}^{-20}\mathrm{J}\right)\end{array}$

Thus, the lowest energy is,

$$\begin{gathered} {\mathrm{E}}_{1,1,1}=3\left(6.0247\times {10}^{-20}\mathrm{J}\right) \\ =1.81\times {10}^{-20}\mathrm{J} \\ =\left(1.81\times {10}^{-20}\mathrm{J}\right)\left(\frac{1\mathrm{eV}}{1.60218\times {10}^{10}\mathrm{J}}\right) \\ =1.13\mathrm{cV} \end{gathered}$$

There is only one such state.

The lowest energy correspond to any of the three choice of quantum numbers,

$$\begin{gathered} \left({\mathrm{n}}_{\mathrm{x}}{\mathrm{n}}_{\mathrm{y}},{\mathrm{n}}_{\mathrm{z}}\right)=\left(2,1,1\right) \\ \left({\mathrm{n}}_{\mathrm{x}}{\mathrm{n}}_{\mathrm{y}},{\mathrm{n}}_{\mathrm{z}}\right)=\left(1,2,1\right) \\ \left({\mathrm{n}}_{\mathrm{x}}{\mathrm{n}}_{\mathrm{y}},{\mathrm{n}}_{\mathrm{z}}\right)=\left(1,1,2\right) \end{gathered}$$

All of which have the same energy as,

$E_{{\mathit{n}}_{\mathit{x}}\mathit{,}{\mathit{n}}_{\mathit{y}}\mathit{,}{\mathit{n}}_{\mathit{z}}}\mathit{=}\mathit{(}{\mathit{n}}_{\mathit{x}}^{\mathit{2}}\mathit{+}{\mathit{n}}_{\mathit{y}}^{\mathit{2}}\mathit{+}{\mathit{n}}_{\mathit{z}}^{\mathit{2}}\mathit{)}\frac{{\mathit{\pi }}^{\mathit{2}}{\mathit{h}}^{\mathit{2}}}{\mathit{2}\mathit{m}{\mathit{L}}^{\mathit{2}}}$

Substitute $1.054\times {10}^{-34}\mathrm{J}.\mathrm{s}$ for h, $9.1094\times {10}^{-31}\mathrm{kg}$ for m, and 1 nm for L in the above equation.

$\begin{array}{r}{\mathrm{E}}_{1,1,2}=\left(1^2+1^2+2^2\right)\frac{{\pi }^2{\mathrm{h}}^2}{2{\mathrm{mL}}^2}\\ =\frac{6(3.14159{)}^2{\left(1.054\times {10}^{-34}\mathrm{J}.\mathrm{s}\right)}^2}{2\left(9.1094\times {10}^{-31}\mathrm{kg}\right)(1\mathrm{nm}{)}^2}\\ =\frac{6(3.14159{)}^2{\left(1.054\times {10}^{-34}\mathrm{J}.\mathrm{s}\right)}^2}{2\left(9.1094\times {10}^{-31}\mathrm{kg}\right)\left(1\mathrm{nm}\right){\left(\frac{{10}^{-9}\mathrm{m}}{\mathrm{nm}}\right)}^2}\\ =6\left(6.0247\times {10}^{-20}\mathrm{J}\right)\\ E_{1,1,2}=\left(3.62\times {10}^{19}\mathrm{J}\right)\left(\frac{\mathrm{J}}{1.6\times {10}^{-10}\mathrm{eV}}\right)\\ =2.26\mathrm{eV}\end{array}$

The lowest energy corresponds to any of the three choice of quantum numbers,

$$\begin{gathered} \left({\mathrm{n}}_{\mathrm{x}}{\mathrm{n}}_{\mathrm{y}},{\mathrm{n}}_{\mathrm{z}}\right)=\left(1,2,2\right) \\ \left({\mathrm{n}}_{\mathrm{x}}{\mathrm{n}}_{\mathrm{y}},{\mathrm{n}}_{\mathrm{z}}\right)=\left(2,1,2\right) \\ \left({\mathrm{n}}_{\mathrm{x}}{\mathrm{n}}_{\mathrm{y}},{\mathrm{n}}_{\mathrm{z}}\right)=\left(2,2,1\right) \end{gathered}$$

All of which have the same energy as,

$E_{n_x\mathit{,}{n}_y\mathit{,}{n}_z}\mathit{=}\mathit{(}{n}_x^{\mathit{2}}\mathit{+}{n}_y^{\mathit{2}}\mathit{+}{n}_z^{\mathit{2}}\mathit{)}\frac{{\pi }^{\mathit{2}}{h}^{\mathit{2}}}{\mathit{2}m{L}^{\mathit{2}}}$$E_{n_x\mathit{,}{n}_y\mathit{,}{n}_z}\mathit{=}\mathit{(}{n}_x^{\mathit{2}}\mathit{+}{n}_y^{\mathit{2}}\mathit{+}{n}_z^{\mathit{2}}\mathit{)}\frac{{\pi }^{\mathit{2}}{h}^{\mathit{2}}}{\mathit{2}m{L}^{\mathit{2}}}$

Substitute $1.054\times {10}^{-34}\mathrm{J}.\mathrm{s}$ for h , $9.1094\times {10}^{-31}\mathrm{kg}$ for m , and 1 nm for L in the above equation.

$\begin{array}{r}{\mathrm{E}}_{1,2,2}=\left(1^2+2^2+2^2\right)\frac{{\pi }^2{\mathrm{h}}^2}{2{\mathrm{mL}}^2}\\ =\frac{9(3.14159{)}^2{\left(1.054\times {10}^{-34}\mathrm{J}.\mathrm{s}\right)}^2}{2\left(9.1094\times {10}^{-31}\mathrm{kg}\right)(1\mathrm{nm}{)}^2}\\ =\frac{9(3.14159{)}^2{\left(1.054\times {10}^{-34}\mathrm{J}.\mathrm{s}\right)}^2}{2\left(9.1094\times {10}^{-31}\mathrm{kg}\right)\left(1\mathrm{nm}\right){\left(\frac{{10}^{-9}\mathrm{m}}{\mathrm{nm}}\right)}^2}\\ =9\left(6.0247\times {10}^{-20}\mathrm{J}\right)\\ {\mathrm{E}}_{1,2,2}=\left(5.42\times {10}^{-19}\mathrm{J}\right)\left(\frac{\mathrm{eV}}{1.6\times {10}^{-19}\mathrm{J}}\right)\\ =3.38\mathrm{eV}\end{array}$

Among the first three energies computed in part (a), there are one ground state, three degenerate first excited states, and three degenerate second excited levels for a total of seven states.