Question

If the Fermi energy of silver is 5.5 eV, the wave number of the fastest electron at 0 K has the magnitude in (meter)^-1 is?

Answer 1

Since, Fermi momentum if

$$P_{ F}=\hbar k_{\small F}$$

Also,

$$P_F=\sqrt{2m_eE_{F}}$$

Thus, from above two equation we get,

$$k_F=\frac{\sqrt{2m_eE_F}}{\hbar}$$

Given,

$$E_F=5.5eV=8.81197^{-19}J\\$$

Thus,

$$k_F\approx 2\times 10^9m^{-1}$$