Show that Eq. 41-9 can be written as ${\mathit{E}}_{\mathbf{F}}\mathbf{=}\mathit{A}{\mathit{n}}^{\mathbf{2}\mathbf{/}\mathbf{3}}$where the constant Ahas the value $\mathbf{3}\mathbf{.}\mathbf{65}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{19}}{\mathbf{m}}^{\mathbf{2}}\mathbf{eV}$.

The equation of Fermi energy according to Eq. 41-9 is

$E_F={\left[\frac{3}{16\sqrt{2}\mathrm{\pi }}\right]}^{2/3}\frac{h^2}{m}{n}^{2/3}...................................\left(1\right)$.

where, $n$is the number density of free electrons, $m\left(=9.1\times {10}^{-31}\mathrm{kg}\right)$ is the electron’s mass and $h\left(=6.63\times {10}^{-34}\mathrm{J}.\mathrm{s}\right)$is the Planck’s constant.

The highest level of the valence band occupied by the electrons at absolute zero temperature is known as Fermi Level. The energy of the electrons occupying the Fermi level is known as Fermi energy.

From the given formula of Fermi energy, we can see that the energy is directly related to the number of conductions electrons having all other terms as constants. Thus, the formula for Femi energy can also be written as-

$E_F=A{n}^{2/3}$

In the above equation, the value of $A$is-

$$\begin{gathered} A={\left[\frac{3}{16\sqrt{2}\mathrm{\pi }}\right]}^{2/3}\frac{h^2}{m} \\ ={\left[\frac{3}{16\sqrt{2}\mathrm{\pi }}\right]}^{2/3}\frac{{\left(6.63\times {10}^{-34}\mathrm{J}.\mathrm{s}\right)}^2}{9.1\times {10}^{-31}\mathrm{kg}} \\ =5.842\times {10}^{-38}{\mathrm{J}}^2.{\mathrm{s}}^2/\mathrm{kg} \end{gathered}$$

Since $1\mathrm{J}=1\mathrm{kg}.{\mathrm{m}}^2/{\mathrm{s}}^2$the units of $A$can be written as ${\mathrm{m}}^2.\mathrm{J}$.