 Solution

Add the electronic diamagnetism of each helium atom to the nuclear magnetic susceptibility, using the same $n_3$ for both contributions in the helium-3 sample:The density cancels when the two terms balance. The sign-change temperature in the independent-atom approximation is thereforeNo numerical atomic radius was supplied, so the exact numerical value remains conditional on that electronic length scale. Taking $⟨r^2⟩\sim a_0^2$, where $a_0$ is the Bohr radius, gives $T_0\simeq 0.32\mathrm{K}$. A slightly more explicit estimate uses the helium variational effective charge $\zeta =27/16$, for which a hydrogenic $1s$ orbital has $⟨r^2⟩=3a_0^2/{\zeta }^2\simeq 1.05a_0^2$, givingAbove this scale the electronic contribution is negative and dominates; below it the positive nuclear contribution dominates within the independent-moment approximation. The estimate is not a precision prediction of a real interacting helium phase.