Answer to Question #240829 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 )
1.The de Broglie wavelength of the electron is ________
2.The Fermi velocity is ________

Answer 1

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

$$v_F = \sqrt{\dfrac{2E_F}{m_e}}$$

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:

$$v_F = \sqrt{\dfrac{2\cdot 1.121\times 10^{-18}}{9.1\times 10^{-31}}} \approx 4.96\times 10^6m/s$$

1. By definition, the de Broglie wavelength of the electron is:

$$\lambda_F = \dfrac{h}{m_e v_F} = \dfrac{h}{\sqrt{2m_eE_F}}\\ \lambda_F = \dfrac{6.6 ×10^{−34}}{\sqrt{2\cdot 1.121\times 10^{-18}\cdot 9.1\times 10^{-31}}} \approx 4.62\times 10^{-10}m$$

Answer. 1) $4.96\times 10^6m/s$, 2) $4.62\times 10^{-10}m$.