Answer
Proven that $A = 3.65 \times 10^{-19} m^2.eV$
Work Step by Step
Fermi energy is given by the equation
$E_F= |\frac{3}{16\sqrt 2} \pi|^{2/3} $ $ \frac{h^2}{m} n^{2/3}$
where
N is number of conduction per unit volume
m is electron mass = $9.109\times10^{-31} kg$
h is Planck Constant = $6.626 \times 10^{-34} j.s$
The N is unknown so equation can be written
$E_F=An^{2/3}$
substitute all the constants. We want to prove A here
$A = |\frac{3}{16\sqrt 2} \pi|^{2/3} $ $ \frac{h^2}{m} $
$A = |\frac{3}{16\sqrt 2} \pi|^{2/3} $ $ \frac{(6.626 \times 10^{-34} j.s)^2}{9.109\times10^{-31} kg} $
$A = 5.842 \times 10^{-38} \frac{J^2 s^2}{kg}$
Since $1 J = 1 \frac{kg m^2}{s^2}$
Dividing by $1.602 \times 10^{-19} J/eV $
yields to $A = 3.65 \times 10^{-19} m^2.eV$
