The Fermi velocity${\mathit{V}}_{\mathbf{f}}$is defined by${\mathit{E}}_{\mathbf{F}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{m}{\mathit{s}}_{\mathbf{F}}^{\mathbf{2}}$, where${\mathit{E}}_{\mathbf{F}}$is the Fermi energy. The Fermi energy for conduction electron in sodium is 3 I$\mathit{V}$. (a) Calculate the Fermis velocity (b) What would be the wavelength of an electron with this velocity? (c) If each sodium atom contributes one conduction electron to the electron gas and sodium atom are spaced roughly$\mathbf{0}\mathbf{.37}\mathbf{}\mathbf{\text{nm}}$apart. If it is necessary, by the criteria of equation (9-43), to treat the conduction electron gas as quantum gas?

The Fermi energy${\mathit{E}}_{\mathbf{F}}$for a particle of mass$\mathit{m}$can be written as

${\mathit{E}}_{\mathbf{F}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{m}{\mathit{v}}_{\mathbf{F}}^{\mathbf{2}}\mathbf{\text{(1)}}$

Where,${\mathit{v}}_{\mathbf{F}}$is Fermi velocity,$\mathit{m}$is mass of electron, which is equal to

$\mathbf{9}\mathbf{.1}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{31}}\mathbf{}\mathbf{\text{kg}}$

The de Broglie wavelength$\mathit{\lambda }$of a particle of mass$\mathit{m}$and velocity$\mathit{v}$is given by

$\mathit{\lambda }\mathbf{=}\frac{\mathbf{h}}{\mathbf{m}\mathbf{v}}\mathbf{\text{(2)}}$

$\mathit{\lambda }\mathbf{=}\frac{\mathbf{h}}{\mathbf{m}\mathbf{v}}\mathbf{\text{(2)}}$

Where,$\mathbf{\text{h}}$is Planck's constant which is equal to$\mathbf{6}\mathbf{.64}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{34}}\mathbf{}\mathbf{\text{J}}\mathbf{\cdot }\mathbf{\text{s}}$

$\mathit{m}$is mass of electron, which is equal to$\mathbf{9}\mathbf{.1}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{31}}\mathbf{}\mathbf{\text{kg}}$

The condition for classical behaviour for a particle of wavelength$\mathit{\lambda }$is

${\mathbf{\left(}\frac{\mathbf{\lambda }}{\mathbf{d}}\mathbf{\right)}}^{\mathbf{3}}\mathbf{<}\mathbf{<}\mathbf{1}\mathbf{\text{(3)}}$

Where,$\mathit{d}$is separation between the particles.

Fermi energy of conduction electrons in sodium is,

$\begin{array}{c}{E}_F=3.1\text{eV}\\ =\left(3.1\text{eV}\right)\left(\frac{1.6\times {10}^{-19}\text{J}}{1\text{eV}}\right)\\ =4.9\times {10}^{-19}\text{J}\end{array}$

Calculation:

Substitute the numerical values in equation (1)

$\begin{array}{c}{v}_F^2=\frac{2\times 4.9\times {10}^{-19}\text{kg}\cdot {\text{m}}^2/{\text{s}}^2}{(9.1\times {10}^{-31}\text{kg})}\\ =1.076\times {10}^{12}{\text{m}}^2/{\text{s}}^2\\ v_F=1.03\times {10}^6\text{m}/\text{s}\end{array}$

Substitute the numerical values in equation (2)

$\begin{array}{c}\lambda =\frac{6.64\times {10}^{-34}{\text{kgm}}^2/\text{s}}{(9.1\times {10}^{-31}\text{kg})(1.03\times {10}^6\text{m}/\text{s})}\\ =0.708\times {10}^{-9}\text{m}\\ =0.708\text{nm}\end{array}$

Separation between the sodium atoms is,$d=$$0.37\text{nm}$

Substitute the numerical values in the left side of equation (3)

$\begin{array}{c}{\left(\frac{\lambda }{d}\right)}^3=\left(\frac{0.708\text{nm}}{0.37\text{nm}}\right)\\ =1.9>1\end{array}$

The value of ${\left(\frac{\lambda }{d}\right)}^3$is greater than 1. Therefore, it is necessary to treat the conduction electron gas as a quantum gas, if each sodium atom contributes one conduction electron to the electron gas and sodium atoms are spaced roughly$0.37\text{nm}$ .