A certain material has a molar mass of 20.0g/mol , Fermi energy of 5.00 eV , and 2 valence electrons per atom. What is the density $\mathbf{(}\mathbf{g}\mathbf{/}{\mathbf{cm}}^{\mathbf{3}}\mathbf{)}$?

1. Molar mass of the material,A = 20 g/mol .
2. Fermi energy of the material,${\mathrm{E}}_{\mathrm{F}}=5\mathrm{eV}$.
3. Each atom has 2 valence electrons.

The number of conduction electrons per unit volume per unit energy range, at a particular energy is given as number density of these conduction electrons.

The equation of Fermi energy is

${\mathbf{E}}_{\mathbf{F}}\mathbf{=}{\left[\frac{3}{16\sqrt{2}\mathrm{\tau \tau }}\right]}^{\mathbf{2}\mathbf{/}\mathbf{3}}\frac{{\mathbf{h}}^{\mathbf{2}}}{\mathbf{m}}{\mathbf{n}}^{\mathbf{2}\mathbf{/}\mathbf{3}}$ (i)

where, n is the number of conduction electrons per unit volume, m is the mass of an electron and h is the Planck’s constant.

The density of a material according to the number of atoms per unit volume and molar mass of a material,

${\mathbf{P}}_{\mathbf{material}}\mathbf{=}{\mathbf{n}}_{\mathbf{atoms}}\mathbf{A}$ (ii)

At first using equation (i) and the given value of Fermi energy, we calculate the number of conduction electrons per unit volume as follows:

$$\begin{gathered} \mathrm{n}=\frac{16\sqrt{2}\mathrm{\pi }}{3}{\left(\frac{{\mathrm{m}}_{\mathrm{e}}{\mathrm{E}}_{\mathrm{F}}}{{\mathrm{h}}^2}\right)}^{3/2} \\ =\frac{16\sqrt{2}\mathrm{\pi }}{3}{\left(\frac{{\mathrm{m}}_{\mathrm{e}}{\mathrm{c}}^2{\mathrm{E}}_{\mathrm{F}}}{{\left(\mathrm{hc}\right)}^2}\right)}^{3/2} \\ =\frac{16\sqrt{2}\mathrm{\pi }}{3}{\left(\frac{\left(0.511\times {10}^6\mathrm{eV}\right)\left(5\mathrm{eV}\right)}{{\left(1240\mathrm{eV}.\mathrm{nm}\right)}^2}\right)}^{3/2}\left(\because {\mathrm{m}}_{\mathrm{e}}{\mathrm{c}}^2=0.511\times {10}^6\mathrm{eV},\mathrm{hc}=1240\mathrm{eV}.\mathrm{nm}\right) \\ =50.9/{\mathrm{nm}}^3 \end{gathered}$$

The number of moles per unit volume is given as-

$$\begin{gathered} \mathrm{n}=\frac{5.09\times {10}^{28}/{\mathrm{m}}^3}{6.022\times {10}^{23}/\mathrm{mol}}\left(\mathrm{Avogadro}\text{'}\mathrm{s}\mathrm{number}=6.022\times {10}^{23}/\mathrm{mol}\right) \\ =8.4\times {10}^4\mathrm{mol}/{\mathrm{m}}^3 \end{gathered}$$

Now, we are given the atoms are bivalent,

$$\begin{gathered} {\mathrm{n}}_{\mathrm{atom}}=\mathrm{n}/2 \\ =\left(8.4\times {10}^4\mathrm{mol}/{\mathrm{m}}^3\right)/2 \\ =4.2\times {10}^4\mathrm{mol}/{\mathrm{m}}^3 \end{gathered}$$

Thus, using this value in equation (ii), we can get the density of the material as follows:

$$\begin{gathered} {\mathrm{P}}_{\mathrm{material}}=\left(4.2\times {10}^4\mathrm{mol}/{\mathrm{m}}^3\right)\left(20\mathrm{g}/\mathrm{mol}\right) \\ =8.4\times {10}^5\mathrm{g}/{\mathrm{m}}^3 \\ =0.84\mathrm{g}/{\mathrm{cm}}^3 \end{gathered}$$

Hence, the value of the density is 0.84 $\mathrm{g}/{\mathrm{cm}}^3$.