Question 5.22: Helium-3 is fermion with spin 1/2 (unlike the more common is...

T_{F}=\frac{E_{F}}{k_{B}}=\frac{\hbar^{2}}{2 m k_{B}}\left(3 \rho \pi^{2}\right)^{2 / 3}   (Equation 5.54).

E_{F}=\frac{\hbar^{2}}{2 m}\left(3 \rho \pi^{2}\right)^{2 / 3}    (5.54).

The number density \rho=\frac{\text { atoms }}{\text { mass }} \times \frac{\text { mass }}{\text { volume }}=\frac{\rho_{m}}{m}, \text { where } \rho_{m} is the mass density and m is the atomic mass-3 times the mass of the proton: m = 3mp (well, OK, two protons and a neutron, and two electrons, and some binding energy . . . but 3 protons is close enough). So

T_{F}=\frac{\hbar^{2}}{6 m_{p} k_{B}}\left(\frac{\pi^{2} \rho_{m}}{m_{p}}\right)^{2 / 3}=\frac{\left(1.055 \times 10^{-34}\right)^{2}}{6\left(1.673 \times 10^{-27}\right)\left(1.381 \times 10^{-23}\right)}\left(\frac{\pi^{2}(82)}{1.673 \times 10^{-27}}\right)^{2 / 3}=4.95 K .