Question

What is the Fermi energy of gold (a monovalent metal with molar mass 197 g/mol and density $\mathbf{19}\mathbf{.}\mathbf{3}\mathbf{}\mathbf{g}\mathbf{/}{\mathbf{cm}}^{\mathbf{3}}$ )?

Short answer

The Fermi energy of gold is $5.52\mathrm{eV}$.

Step-by-step solution

The given data

• Molar mass of gold, A = 197 g/mol
• Density of gold,$d=19.30g/{\mathrm{cm}}^3$
• Gold is a monovalent metal.

Understanding the concept of Fermi energy

Using the given formula for the number of atoms per unit volume, we can get the required value of the conduction electrons per unit volume, considering that metal is monovalent. Now using this value in the equation of Fermi energy, we can calculate the Fermi energy of the metal.

Formulae:

The mass of an atom, $\mathbf{M}\mathbf{=}\mathbf{A}\mathbf{/}{\mathbf{N}}_{\mathbf{A}}\mathbf{,}\mathbf{}\mathbf{where}\mathbf{}{\mathbf{N}}_{\mathbf{A}}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{022}\mathbf{\times }{\mathbf{10}}^{\mathbf{23}}\mathbf{}{\mathbf{mol}}^{\mathbf{-}\mathbf{1}}$ (i)

The number density of conduction electrons, $\mathbf{n}\mathbf{=}\frac{\mathbf{d}}{\mathbf{M}}$ (ii)

d=density of the atom, M = mass of a single atom

The equation of Fermi energy is ${\mathit{E}}_{\mathbf{F}}\mathbf{=}\frac{\mathbf{0}\mathbf{.}\mathbf{121}\mathbf{}{\mathbf{h}}^{\mathbf{2}}}{\mathbf{m}}{\mathit{n}}^{\mathbf{2}\mathbf{/}\mathbf{3}}$ (iii)

Calculation of Fermi energy

As gold is monovalent metal, each atom contributes one conduction electron; thus, the number density of conduction electrons should be equal to the number density of atoms.

For the given data, comparing equation (i) and (ii), we get the number density of the conduction electrons in gold as follows:

$$\begin{gathered} n=\frac{d}{A/N_A} \\ =\frac{\left(19.3g/c{m}^3\right)}{\left(197g/mol\right)/\left(6.022\times {10}^{23}mo{l}^{-1}\right)} \\ =5.90\times {10}^{22}c{m}^{-3} \\ =59.0n{m}^{-3} \end{gathered}$$

Using the above number density value in equation (iii), the Fermi energy of gold metal can be calculated as follows:

$$\begin{gathered} E_F=\frac{0.121{\left(hc\right)}^2}{\left(m_e{c}^2\right)}{n}^{2/3} \\ =\frac{0.121{\left(1240eV.nm\right)}^2}{511\times {10}^{3}eV}{\left(59.0n{m}^{-3}\right)}^{2/3} \\ =5.52eV \end{gathered}$$

Hence, the value of Fermi energy is 5.52 eV.