= Solution
A spin-half gas has two spin states. Filling the <Fermi sphere> at zero <temperature> gives $n=k_F^3/(3\pi^2)$, so
$$
k_F=(3\pi^2\rho/m_3)^{1/3},\qquad
E_F=\frac{\hbar^2k_F^2}{2m_3},\qquad v_F=\frac{\hbar k_F}{m_3}.
$$
With $m_3\simeq5.008\times10^{-27}\,\mathrm{kg}$, the number density is $1.62\times10^{28}\,\mathrm{m^{-3}}$ and $k_F\simeq7.82\times10^9\,\mathrm{m^{-1}}$. Thus
$$
\boxed{E_F\simeq6.80\times10^{-23}\,\mathrm J\simeq0.424\,\mathrm{meV},\qquad
v_F\simeq165\,\mathrm{m\,s^{-1}}.}
$$
The associated <Fermi temperature> is
$$
\boxed{T_F=E_F/k_B\simeq4.92\,\mathrm K.}
$$
Fermionic degeneracy matters on cooling to a few kelvin or below in this model, and is pronounced for $T\ll T_F$. Liquid helium-3 is interacting; its quasiparticle effective <mass> and thermodynamics need not equal the bare-mass free-gas predictions. The requested calculation is the ideal Fermi-gas scale.
Back to article page
