For a particle in a one-dimensional (ID) box, ${\mathbf{E}}_{\mathbf{n}}$is proportional to a single quantum number $\mathbf{n}$. Let us simplify things by ignoring the proportionality factor: ${\mathbf{E}}_{\mathbf{n}}\mathbf{=}{\mathbf{n}}^{\mathbf{2}}$ . For a 3D box, ${\mathbf{E}}_{{\mathbf{n}}_{\mathbf{x}}\mathbf{,}{\mathbf{n}}_{\mathbf{y}}\mathbf{,}{\mathbf{n}}_{\mathbf{z}}}\mathbf{=}{\mathbf{n}}_{\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{n}}_{\mathbf{y}}^{\mathbf{2}}\mathbf{+}{\mathbf{n}}_{\mathbf{z}}^{\mathbf{2}}$, and the 2D box is fairly obvious.

(a) The table shows a start on accounting for allowed states. Complete the table, stopping after the 10th state (state, not energy) for all three cases.

(b) Find the number of states per energy difference for the first five states and the last five states for all three cases. For instance, for the first five in the ID case, it is 5 states per energy difference of 24, or$\mathbf{5}\mathbf{/}\mathbf{24}$ .

(c) Overlooking the obviously crude aspects of this accounting, does the "density of states" seem to increase with energy, decrease with energy, or stay about the same?

The expression of number of states is given by,

$\mathbf{N}\mathbf{=}\frac{\mathbf{n}}{{\mathbf{E}}_{\mathbf{2}}\mathbf{-}{\mathbf{E}}_{\mathbf{1}}}$

a)

Given information is

The ${\mathrm{E}}_{\mathrm{n}}$ is proportional to single quantum number $n$.

To complete the table up to ${10}^{\text{th}}$ state, $\mathrm{n}$ should varies such that it gives the lowest amount of energy. The complete table is

(b)

The expression of number of states is given by,

$\mathrm{N}=\frac{\mathrm{n}}{{\mathrm{E}}_2-{\mathrm{E}}_1}$

The number of states for first five in 1D is calculated as,

$\begin{array}{c}\mathrm{N}=\frac{\mathrm{n}}{{\mathrm{E}}_2-{\mathrm{E}}_1}\\ \mathrm{N}\left(\text{firstfive}\right)=\frac{5}{25-1}\\ =\frac{5}{24}\end{array}$

The number of states for last five in 1Dis calculated as,

$\begin{array}{c}\mathrm{N}=\frac{\mathrm{n}}{{\mathrm{E}}_2-{\mathrm{E}}_1}\\ \mathrm{N}\left(\text{lastfive}\right)=\frac{5}{100-36}\\ =\frac{5}{64}\end{array}$

The number of states for first five in 2Dis calculated as,

$\mathrm{N}=\frac{\mathrm{n}}{{\mathrm{E}}_2-{\mathrm{E}}_1}$

$\begin{array}{c}\mathrm{N}\left(\text{firstfive}\right)=\frac{5}{10-2}\\ =\frac{5}{8}\end{array}$

The number of states for last five in 2Dis calculated as,

$\begin{array}{c}\mathrm{N}=\frac{\mathrm{n}}{{\mathrm{E}}_2-{\mathrm{E}}_1}\\ \mathrm{N}\left(\text{lastfive}\right)=\frac{5}{17-10}\\ =\frac{5}{7}\end{array}$

The number of states for first five in 3Dis calculated as,

$\begin{array}{c}\mathrm{N}=\frac{\mathrm{n}}{{\mathrm{E}}_2-{\mathrm{E}}_1}\\ \mathrm{N}\left(\text{firstfive}\right)=\frac{5}{9-3}\\ =\frac{5}{6}\end{array}$

The number of states for last five in 3Dis calculated as,

$\begin{array}{c}\mathrm{N}=\frac{\mathrm{n}}{{\mathrm{E}}_2-{\mathrm{E}}_1}\\ \mathrm{N}\left(\text{lastfive}\right)=\frac{5}{11-9}\\ =\frac{5}{2}\end{array}$

Therefore, the number of states for first five and last five in $1D$ are $\frac{5}{24}$ and $\frac{5}{64}$ respectively, in 2D are $\frac{5}{8}$ and $\frac{5}{7}$ respectively and in 3D are $\frac{5}{6}$ and $\frac{5}{2}$ respectively.

c)

From part (b), the density of the states tends to increase with the dimensionality but decrease with the energy.