Answer
$E_F = 5.52 $
Work Step by Step
Use the simplified equation of number density conduction equation
Where molar mass of gold $ A = 197 g/mol$
$n = \frac{(19.3 g/cm^3)(6.022 \times 10^{23} /mol)}{197 g/mol} $
$n = 5.90 \times 10^{28} m^{-3}$
To find Fermi Energy,
$E_F = \frac{(0.121)(hc)^2}{m_e c^{2}} n^{2/3}$
where
$hc = 1240 eV.nm$
$m_e c^2 = 511 \times 10^3 eV$
$n = 5.90 \times 10^{28} m^{-3}$
Substitute into equation
$E_F = \frac{(0.121)(1240 eV.nm^3)}{(511 \times 10^3 eV)} (5.90 \times 10^{28} m^{-3})^{2/3}$
$E_F = 5.52 $
