Question

A divalent bcc solid has a lattice constant of 4.5 Å. Calculate its Fermi energy

Answer 1

As per the given question,

Lattice constant =$4.5A^o$

Fermi energy =?

We know that electron density $n=\dfrac{2}{a^3}=\dfrac{2}{(4.5\times 10^{-10})^3}=0.0219\times 10^{30}/m^3$

Fermi wave factor $K_F=(3\pi^2 n)^{1/3}=(3\pi^2 0.0219\times 10^{30})^{1/3}=0.86\times 10^{10} /m$

Fermi energy =$\dfrac{n^2K_{F}^2}{2m}=\dfrac{(0.0219\times 10^{30})^2\times (0.86\times10^{10})^2}{2\times 9.11\times 10^{-31}}$

$=\dfrac{3.547\times 10^{76}}{18.22\times 10^{-31}}=0.19\times 10^{107}J$