Question

A box of volume 1 cm³ contains 5.2×10²¹ electrons. Calculate their Fermi

momentum and Fermi energy. Take m_e=9.1*10^-28g ,m_n=1.67×10^-24g 1.67×10^-27 ergs.

Answer 1

The Fermy energy can be calculated as follows:

$$E_F=\dfrac{h^2}{2m_e}(\dfrac{3N}{8\pi V})^{2/3},$$ $$E_F=\dfrac{(6.62\cdot10^{-34}\ J\cdot s)^2}{2\cdot9.1\cdot10^{-31}\ kg}\cdot(\dfrac{3\cdot5.2\cdot10^{21}\ el}{8\pi\cdot1\cdot10^{-6}\ m^3})^{2/3}=1.75\cdot10^{-19}\ J.$$

By the definition of the Fermi momentum, we have:

$$p_F=\sqrt{2m_eE_F}=\sqrt{2\cdot9.1\cdot10^{-31}\ kg\cdot1.75\cdot10^{-19}\ J}=5.64\cdot10^{-25}\ \dfrac{kgm}{s}.$$