91Ó°ÊÓ

We have given,

Every atom in a chunk of copper is contributes one conduction electron.

We have to find the Fermi energy, Fermi temperature and pressure.

Formula of the Fermi energy is given by,

${\mathrm{Î\mu }}_{\mathrm{f}}=\frac{{\mathrm{h}}^2}{8\mathrm{m}}{\left(\frac{3\mathrm{N}}{\mathrm{Ï€³Õ}}\right)}^{\frac{2}{3}}$

Where, V is volume of the chunk, which is given by,

$\mathrm{V}=\frac{\mathrm{M}}{\mathrm{ÒÏ}}$

Where we know that the one mole of mass of copper is $63.5\mathrm{g}$and density of the copper is $8.93\mathrm{g}/{\mathrm{cm}}^3.$

Then volume will be,

$$\begin{gathered} \mathrm{V}=\frac{63.5\mathrm{g}}{8.93\mathrm{g}/{\mathrm{cm}}^3} \\ \mathrm{V}=7.1{\mathrm{cm}}^3 \\ \mathrm{V}=7.1×{10}^{-6}{\mathrm{m}}^3 \end{gathered}$$

Since it is given that the every atom is give the one electron. then, number of electron for per unit volume will be equal to number of atom in one mole.

Then,

$\mathrm{N}=6.022×{10}^{23}\mathrm{electron}/\mathrm{mole}$

Mass of the electron =$9.1×{10}^{-31}\mathrm{kg}$

Then, the Fermi energy will be,

$$\begin{gathered} {\mathrm{Î\mu }}_{\mathrm{f}}=\frac{{\mathrm{h}}^2}{8\mathrm{m}}{\left(\frac{3\mathrm{N}}{\mathrm{Ï€³Õ}}\right)}^{\frac{2}{3}} \\ {\mathrm{Î\mu }}_{\mathrm{f}}=\frac{{(6.63×{10}^{-34}\mathrm{J}.\mathrm{s})}^2}{8(9.1×{10}^{-31}\mathrm{kg})}\left(\frac{3(6.022×{10}^{23}}{\mathrm{Ï€}×7.1×{10}^{-6}}\right) \\ {\mathrm{Î\mu }}_{\mathrm{f}}=7.04\mathrm{eV} \end{gathered}$$

Then, the Fermi temperature will be found out by,

${\mathrm{Î\mu }}_{\mathrm{f}}={\mathrm{K}}_{\mathrm{B}}\mathrm{T}$

Where ${\mathrm{K}}_{\mathrm{B}}$is Boltzmann's constant.

then,

localid="1649951480186" $$\begin{gathered} {\mathrm{Î\mu }}_{\mathrm{f}}=(1.38×{10}^{-23}\mathrm{J}/\mathrm{K})\mathrm{T} \\ \mathrm{T}=\frac{1.13×{10}^{-18}\mathrm{J}}{1.38×{10}^{-23}\mathrm{J}/\mathrm{K}} \\ \mathrm{T}=81884\mathrm{K} \end{gathered}$$

The degeneracy pressure can be found as

$$\begin{gathered} \mathrm{P}=\frac{2\mathrm{N}}{5\mathrm{V}}{\mathrm{Î\mu }}_{\mathrm{f}} \\ \mathrm{P}=\frac{2×6.022×{10}^{23}}{5×7.1×{10}^{-6}{\mathrm{m}}^3}(1.13×{10}^{-18}\mathrm{J}) \\ \mathrm{P}=3.74\mathrm{pascal} \end{gathered}$$

The contribution of the degeneracy pressure to the bulk modulus is given by,

$$\begin{gathered} \mathrm{B}=\frac{2\mathrm{N}}{3\mathrm{V}}{\mathrm{Î\mu }}_{\mathrm{f}} \\ \mathrm{B}=\frac{2×6.022×{10}^{23}}{3×7.1×{10}^{-6}{\mathrm{m}}^3}×1.13×{10}^{-18}\mathrm{J} \\ \mathrm{B}=6.3×{10}^5\mathrm{atm} \end{gathered}$$

The found temperature is very large than the room temperature that is why it is sufficiently low to treat this system as a degenerate electron gas.