Question

Question: The Fermi velocity V_F is defined by${\mathbf{E}}_{\mathbf{F}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{mv}}_{\mathbf{F}}^{\mathbf{2}}$ , where is the fermi energy. The Fermi energy for silver is 5.5eV.(a) Calculate the Fermi velocity.(b) what would be the wavelength of an electron with this velocity. (c)How does this compare with the lattice spacing of 0.41 nm? Does the order of magnitude makes sence?

Short answer

Answer

1. The fermi velocity corresponding to the given fermi energy is$1.39\times {\text{10}}^{6}m/s$.
2. The wavelength corresponding to the fermi velocity is$5.23\times {10}^{-10}\text{m}$.
3. The wavelength is of the same order as the lattice vibrations, which is a sensible result.

Step-by-step solution

Given Data

Fermi velocity (V_F)is obtained from the formula $E_F=\frac{1}{2}m{v}_F^2····························\left(1\right)$.

The Fermi energy (E_F) of silver is 5.5eV.

Concept of Fermi Energy

The highest energy level that is occupied by electrons at the absolute zero temperature, is called the Fermi energy level. At this temperature, the fermi level lies in the middle of the valence band and the conduction band.

Calculations

(a) Using equation (1) for$E_F=5.5\text{eV}$, we get

$$\begin{gathered} E_F=\frac{1}{2}m{v}_F^2 \\ {v}_F=\sqrt{\frac{2E_F}{m}} \end{gathered}$$

Here, m is the mass of electron.

For the given values, the above equation becomes-

$$\begin{gathered} v_F=\sqrt{\frac{2E_F}{m}} \\ =\sqrt{\frac{2\times 5.5\text{eV}\times \text{1.6}\times {\text{10}}^{-19}\text{J}}{9.1\times {\text{10}}^{-31}\text{kg}\times \text{1}\text{eV}}} \\ =1.39\times {\text{10}}^6\text{m/s} \end{gathered}$$

The fermi velocity is$1.39\times {\text{10}}^{6}m/s$.

(b) the wavelength$\lambda$ corresponding to a given energy is given as-

$\lambda =\frac{h}{mv}················································\left(2\right)$

For, $v_F=1.39\times {\text{10}}^{6}m/s$the above equation becomes-

$$\begin{gathered} \lambda =\frac{\left(6.626\times {10}^{-34}\text{Js}\right)}{\left(9.1\times {10}^{-31}\text{kg}\right)\left(1.39\times {10}^{6}m/s\right)} \\ =5.23\times {10}^{-10}\text{m} \end{gathered}$$

The wavelength corresponding to the Fermi velocity is$5.23\times {10}^{-10}\text{m}$.

(c) the wavelength is certainly of the same order as the lattice spacing, which is sensible for the electrons undergoing transition between the bands.