91Ó°ÊÓ

At T = 0, equation 8.50 says that$\stackrel{¯}{s}=1$ . Work out the first temperature-dependent correction to this value, in the limit $\mathrm{β}∈n≫1$.

We must change the Fermi energy since each spatial wave function may carry four nucleons, hence the first equation 7.38 must be multiplied by a factor of two, yielding:

$N=\frac{2\mathrm{Ï€}{n}_{\max }^3}{3}$

Solve for n_max:

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

Fermi energy in terms of n_max:

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

Substitute with n_max:

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

Where, V=L³

The number density of gas is:

$\frac{N}{V}=0.18{\mathrm{fm}}^{-3}=0.18\frac{1}{{\mathrm{fm}}^3}×\frac{{\mathrm{fm}}^3}{{\left(1.0×{10}^{-15}\right)}^3{\mathrm{m}}^3}=1.8×{10}^{44}{\mathrm{m}}^{-3}$

Substitute with the values:

$$\begin{gathered} {\mathrm{Ï\mu }}_F=\frac{{\left(6.626×{10}^{-34}\mathrm{J}·\mathrm{s}\right)}^2}{8\left(1.67×{10}^{-27}\mathrm{kg}\right)}{\left(\frac{3\left(1.8×{10}^{44}{\mathrm{m}}^{-3}\right)}{2\mathrm{Ï€}}\right)}^{2/3} \\ =6.40×{10}^{-12}\mathrm{J} \\ {\mathrm{Ï\mu }}_{\mathrm{F}}=40\mathrm{MeV} \end{gathered}$$

The Fermi energy is calculated by multiplying the Boltzmann constant by the Fermi temperature, which is:

$$\begin{gathered} {\mathrm{Ï\mu }}_F=k{T}_F \\ {T}_F=\frac{{\mathrm{Ï\mu }}_F}{k} \end{gathered}$$

Substitute with fermi energy:

$$\begin{gathered} T_F=\frac{6.40×{10}^{-12}\mathrm{J}}{1.38×{10}^{-23}\mathrm{J}/\mathrm{K}} \\ {T}_F=4.638×{10}^{11}\mathrm{K} \end{gathered}$$

This is hotter than the centre of any ordinary star. We can treat the nucleus as a degenerate.