Calculate the density of states $\mathit{N}\mathbf{\left(}\mathit{E}\mathbf{\right)}$for metal at energy $\mathit{E}\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{0}\mathit{e}\mathit{V}$and show that your result is consistent with the curve of Fig. 41-6.

Energy of the metal,$E=8\mathrm{eV}$

The number of states per unit energy range per unit volume $\left(N\left(E\right)\right)$, present in a sample of the material at a particular energy $\left(E\right)$, is known as density of states. The formula for density of states is given as-

$$\begin{gathered} N\left(E\right)=\frac{8\sqrt{2}{\mathrm{\pi m}}^{3/2}}{h^3}{E}^{1/2}.............................\left(1\right) \\ \mathrm{where}\mathrm{h}=6.63\times {10}^{-34}\mathrm{J}.\mathrm{s}\mathrm{and}\mathrm{m}=9.1\times {10}^{-31}\mathrm{kg} \end{gathered}$$

We can write equation (1) as follows:

$N\left(E\right)=C{E}^{1/2}$

In the above equation, the value of C is -

$$\begin{gathered} C=\frac{8\sqrt{2}{\mathrm{\pi m}}^{3/2}}{h^3} \\ =\frac{8\sqrt{2}\mathrm{\pi }{\left(9.1\times {10}^{-31}\mathrm{kg}\right)}^{3/2}}{\left(6.63\times {10}^{-34}\mathrm{J}.\mathrm{s}\right)} \\ =1.062\times {10}^{56}{\mathrm{kg}}^{3/2}/{\mathrm{J}}^3.{\mathrm{s}}^3 \\ =6.81\times {10}^{27}{\mathrm{m}}^{-3}.{\left(\mathrm{eV}\right)}^{-2/3} \end{gathered}$$

Using the given data in equation (1), the density of states for the metal with energy $E=8eV$can be calculated as follows:

$$\begin{gathered} N\left(E\right)=\left[6.81\times {10}^{27}{\mathrm{m}}^{-3}.{\left(\mathrm{eV}\right)}^{-2/3}\right]{\left(8\mathrm{eV}\right)}^{1/2} \\ =1.9\times {10}^{28}{\mathrm{m}}^{-3}.{\left(\mathrm{eV}\right)}^{-1} \end{gathered}$$

This value of density of state is consistent with the given figure 41-6.

Hence, the value of the density of states is $1.9\times {10}^{28}{\mathrm{m}}^{-3}.{\left(\mathrm{eV}\right)}^{-1}$.