Question

What is the Fermi energy of gold (a monovalent metal with molar mass $$197g/mol$$ and density $$19.3g/c{m}^3$$ ). Given $$hc=1240eVnm$$

Answer 1

Hint: We are asked to find the value of Fermi energy, so we write the appropriate formula but we see that the quantity of number density is missing. Then we move on to utilizing the density given to find the number density. Now that we have the number density, we can find the Fermi energy easily.

Formulas used:
The number density can be found by the formula,
$$n={\displaystyle \frac{\rho N_A}{m}}$$
The fermi energy can be found using the formula,
$$E_F={\displaystyle \frac{0.121h^2}{m_e}}{\left(n\right)}^{2/3}$$
Where $$m_e$$ is the mass of an electron, $$h$$ is the Planck’s constant, $$\rho$$ is the density of gold, $$N_A$$ is the Avogadro number and $$m$$ is the molar mass of gold.

Complete answer:
The following data is given to us,
The molar mass of the gold atom is, $$m=197g/mol$$
The density of gold is, $$\rho =19.3g/c{m}^{-3}$$
For calculating easily it is given that, $$hc=1240eVnm$$
We are asked to find the value of fermi energy of the given monovalent gold atom. Let us start by writing down the formula to find the fermi energy. The fermi energy can be found using the formula,
$$E_F={\displaystyle \frac{0.121h^2}{m_e}}{\left(n\right)}^{2/3}$$
Here we can see that the value of energy density is not with us. So, we move on to find the number density. It is given by,
$$n={\displaystyle \frac{\rho N_A}{m}}$$

Substituting the values that we have,
$$\begin{gathered} n={\displaystyle \frac{\rho N_A}{m}} \\ \Rightarrow n={\displaystyle \frac{19.3\times {10}^4\times 6.022\times {10}^{23}}{0.197}} \\ \Rightarrow n=5.90\times {10}^{28}{m}^{-3} \end{gathered}$$
Since the values are not in the SI units, we convert it to get the above value.Now that we have the value of number density, we can find the value of Fermi energy using the formula given above.
$$\begin{gathered} E_F={\displaystyle \frac{0.121h^2}{m_e}}{\left(n\right)}^{2/3} \\ \Rightarrow E_F={\displaystyle \frac{0.121\times \left(6.626\times {10}^{-34}\right)}{9.1\times {10}^{-31}}}{\left(5.9\times {10}^{28}\right)}^{2/3} \\ \therefore E_F=5.52eV \end{gathered}$$

Therefore, the Fermi energy of gold is, $$5.52eV$$.

Note: The number density is a quantity used to portray the level of grouping of countable items in actual space: three-dimensional volumetric number density, two-dimensional areal number density, or one-dimensional direct number density. Fermi energy is the energy difference between the highest and lowest occupied single-particle states in a quantum system.