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The energy of fermions at absolute zero temperature is known as Fermi energy. The mathematical equation for the Fermi energy is,

$E_F=\frac{{蟺}^2{\mathrm{鈩�}}^2}{m}{\left[\frac{3}{(2s+1)蟺\sqrt{2}}\frac{N}{V}\right]}^{\frac{2}{3}}$ 鈥︹��. (1)

Here,

$\text{N}鈫�$Number of oscillators

$\mathrm{鈩�}鈫�$Planck's reduced constant

$V鈫�$Volume

$s鈫�$Spin

Density of copper$=8.9脳{10}^3\text{kg}/{\text{m}}^3$

The NN in the equation (1) can be rewritten as the mass per unit volume over the mass per atom, or the bulk density D over the atomic mass:

role="math" localid="1658381923377" $\begin{array}{c}{E}_F=\frac{{蟺}^2{\mathrm{鈩�}}^2}{m}{\left[\frac{3}{(2s+1)蟺\sqrt{2}}\frac{N}{V}\right]}^{\frac{2}{3}}\\ =\frac{{蟺}^2{\mathrm{鈩�}}^2}{m}{\left[\frac{3}{(2s+1)蟺\sqrt{2}}\frac{D}{m_A}\right]}^{\frac{2}{3}}\end{array}$

Substitute$9.1脳{10}^{鈭�31}\text{kg}$for mass of the electron $\left(\text{m}\right),8.9脳{10}^3\text{kg}/{\text{m}}^3$for density of copper(D), $1.055脳{10}^{鈭�34}\text{J}.s$for $\mathrm{鈩�},\frac{1}{2}$forthe spin of the electron and $1.055脳{10}^{鈭�25}\text{kg}$for$m_A$.

role="math" localid="1658381931536" $\begin{array}{c}{E}_F=\frac{{蟺}^2{(1.055脳{10}^{鈭�34}\text{J}.\text{s})}^2}{(9.1脳{10}^{鈭�31}\text{kg})}{\left[\frac{3}{(2\left(\frac{1}{2}\right)+1)蟺\sqrt{2}}\frac{(8.9脳{10}^3\text{kg}/{\text{m}}^3)}{(1.055脳{10}^{鈭�25}\text{kg})}\right]}^{\frac{2}{3}}\\ =1.126脳{10}^{鈭�18}\text{J}\end{array}$

Convert the unit for Fermi energy for copper energy from $J$to$\text{eV}$ .

role="math" localid="1658381943480" $\begin{array}{c}{E}_F=1.126脳{10}^{鈭�18}J\\ =(1.126脳{10}^{鈭�18}J)\left(\frac{1\text{eV}}{1.6脳{10}^{鈭�9}J}\right)\\ =7.04\text{eV}\end{array}$

Therefore, the Fermi energy for the copper is$7.04\text{eV}$ .

Since the energy that a particle at room temperature would have been $0.04\text{eV}$(from $\frac{3}{2}{k}_{B}T$), as opposed to the Fermi energy of $7.0\text{eV}$, room temperature would be considered cold in comparison.