Answer to Question #240828 in Physics for Sri

Question

The Fermi enrgy of copper is 7 eV. (use h = 6.6 ×10−34JHz−1 , 1eV = 1.602×10−19J, me=9.1×10−31 kg )
The Fermi momentum of electron in copper is ________

16.48×10−25 kg m/s


13.38×10−25 kg m/s


12.48×10−25 kg m/s


14.28×10−25 kg m/s

Answer 1

The Fermi momentum is given as follows (see https://en.wikipedia.org/wiki/Fermi_energy):

$$p_F = \sqrt{2E_Fm_e}$$

where $E_F = 7eV = 7\cdot 1.602×10^{−19}J \approx 1.121\times 10^{-18}J$ is the Fermi energy, and $m_e = 9.1×10^{−31} kg$ is the mass of electron. Thus, obtain:

$$p_F = \sqrt{2\cdot 1.121\times 10^{-18}\cdot 9.1×10^{−31}} \approx 14.28\times 10^{-25}kg\cdot m/s$$

Answer. 14.28×10−25 kg m/s.