91Ó°ÊÓ

We are given that the number density of gas of nucleons is $0.18{\mathrm{fm}}^{-3}$.

We have to find the fermi energy of given system and the Fermi temperature.

An atomic nucleus can be modeled as a gas of nucleons with number density,

$\left(\frac{N}{V}\right)=0.18{\mathrm{fm}}^{-3}$

Since the nucleons are fourfold degenerate, then the total number of occupied states will be,

$$\begin{gathered} N=4×\left(\text{Volume of eigth-sphere}\right) \\ =4\frac{1}{8}\left(\frac{4}{3}\mathrm{π}{n}_{\max }^3\right) \\ =\frac{2}{3}\mathrm{π}{n}_{\max }^3 \end{gathered}$$

Rearranging the equation for $n_{\max }$, we get

$$\begin{gathered} \frac{3N}{2\mathrm{Ï€}}=n_{\max }^3 \\ {n}_{\max }={\left(\frac{3N}{2\mathrm{Ï€}}\right)}^{\frac{1}{3}} \end{gathered}$$

The fermi energy of the system is given by,

${\mathrm{Î\mu }}_F=\frac{h^2{n}_{\max }^2}{8m{L}^2}$

Putting $n_{max}={\left(\frac{3N}{2\mathrm{Ï€}}\right)}^{\frac{1}{3}}$, we get

$$\begin{gathered} {\mathrm{Î\mu }}_F=\frac{h^2}{8m{L}^2}{\left(\frac{3N}{2\mathrm{Ï€}}\right)}^{\frac{2}{3}} \\ =\frac{h^2}{8m{\left(L^3\right)}^{\frac{2}{3}}}{\left(\frac{3N}{2\mathrm{Ï€}}\right)}^{\frac{2}{3}} \\ =\frac{h^2}{8m}{\left(\frac{3}{2\mathrm{Ï€}}·\frac{N}{V}\right)}^{\frac{2}{3}} \end{gathered}$$

The mass of nucleon is,

$m=\frac{1\mathrm{g}}{N_A}$

Here, $N_A$is Avogadro's number.

Substituting $6.023×{10}^{23}\text{atoms}/\mathrm{mol}$, we get

$$\begin{gathered} m=\frac{(1\mathrm{g})(1\mathrm{kg}/1000\mathrm{g})}{6.023×{10}^{23}} \\ =1.6603×{10}^{-27}\mathrm{kg} \end{gathered}$$

The number density of nucleons in the gas is $0.18{\mathrm{fm}}^{-3}$. Converting number density from fm to $m^{-3}$.

$$\begin{gathered} \frac{N}{V}=\left(0.18{\mathrm{fm}}^{-3}\right){\left(\frac{{10}^{-15}\mathrm{m}}{\mathrm{fm}}\right)}^{-3} \\ =0.18×{10}^{45}{\mathrm{m}}^{-3} \end{gathered}$$

Substituting the values, we get

localid="1647858933038" $$\begin{gathered} {\mathrm{Î\mu }}_F=\frac{{\left(6.625×{10}^{-34}\mathrm{J}·\mathrm{s}\right)}^2}{8\left(1.6603×{10}^{-27}\mathrm{kg}\right)}{\left(\frac{3}{2\mathrm{Ï€}}\right)}^{\frac{2}{3}}{\left(0.18×{10}^{45}{\mathrm{m}}^{-3}\right)}^{2/3} \\ =\left(6.4355×{10}^{-12}\mathrm{J}\right)\left(\frac{1\mathrm{eV}}{1.6×{10}^{-19}\mathrm{J}}\right) \\ =\left(4.0223×{10}^7\mathrm{eV}\right)\left(\frac{1\mathrm{MeV}}{{10}^6\mathrm{eV}}\right) \\ =40.2\mathrm{MeV} \end{gathered}$$

Hence, the Fermi energy of the system is$40.2\mathrm{MeV}$.

The Fermi temperature of the system is given by,

$\left(T_F\right)=\frac{E_F}{k_B}$

Putting ${\mathrm{Î\mu }}_F=4.0223×{10}^7\mathrm{eV}$and

$k_B=8.617×{10}^{-5}\mathrm{eV}/\mathrm{K}$, we get

$$\begin{gathered} T_F=\frac{4.0223×{10}^7\mathrm{eV}}{8.617×{10}^{-5}\mathrm{eV}/\mathrm{K}} \\ =4.6632×{10}^{11}\mathrm{K} \end{gathered}$$

Hence, the Fermi temperature of the system is$4.6632×{10}^{11}\mathrm{K}$and lot of energy is required to excite those nucleons.