= Solution
Use a three-dimensional <ideal gas> at the same <number density> n for both <isotopes>. An unpolarized <helium-3> gas has <spin degeneracy> two, hence
$$
k_BT_F=\frac{\hbar^2}{2m_3}(3\pi^2n)^{2/3}.
$$
<Helium-4> has a single <spin> state, and the ideal Bose-gas saturation condition gives
$$
k_BT_B=\frac{2\pi\hbar^2}{m_4}\left[\frac{n}{\zeta(3/2)}\right]^{2/3}.
$$
Eliminating n, using $m_3/m_4\simeq3/4$ and $\zeta(3/2)=2.612$, yields
$$
\boxed{\frac{T_B}{T_F}=\frac{4\pi}{(3\pi^2)^{2/3}\zeta(3/2)^{2/3}}\frac{m_3}{m_4}
\simeq0.519,\qquad T_B\simeq2.6\,\mathrm K.}
$$
The fermionic <spin degeneracy> must be included. This is an ideal-gas estimate; strong interactions in liquid helium prevent identifying it with an exact experimental transition <temperature>.
Back to article page
