An atomic nucleus can be crudely modeled as a gas of nucleons with a number density of $0.18{\mathrm{fm}}^{-3}$(where $1\mathrm{fm}={10}^{-15}\mathrm{m}$). Because nucleons come in two different types (protons and neutrons), each with spin $1/2$, each spatial wavefunction can hold four nucleons. Calculate the Fermi energy of this system, in MeV. Also calculate the Fermi temperature, and comment on the result.

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,

$$\begin{gathered} N=4\times \left(\text{Volume of eigth-sphere}\right) \\ =4\frac{1}{8}\left(\frac{4}{3}\pi n_{\max }^3\right) \\ =\frac{2}{3}\pi n_{\max }^3 \end{gathered}$$

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

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

The fermi energy of the system is given by,

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

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

$$\begin{gathered} {\epsilon }_F=\frac{h^2}{8m{L}^2}{\left(\frac{3N}{2\pi }\right)}^{\frac{2}{3}} \\ =\frac{h^2}{8m{\left(L^3\right)}^{\frac{2}{3}}}{\left(\frac{3N}{2\pi }\right)}^{\frac{2}{3}} \\ =\frac{h^2}{8m}{\left(\frac{3}{2\pi }·\frac{N}{V}\right)}^{\frac{2}{3}} \end{gathered}$$

The mass of nucleon is,

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

Here, $N_A$is Avogadro's number.

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

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

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

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

Substituting the values, we get

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

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 ${\epsilon }_F=4.0223\times {10}^7\mathrm{eV}$and

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

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

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