 Solution

The helium atom has a closed $1s^2$ electronic shell with total electronic spin and orbital angular momentum zero. The nucleus of helium-4 also has spin zero. At low temperature there is no Curie response from permanent atomic or nuclear moments; the leading atomic response is diamagnetism.

Let $n_4$ denote the number of atoms per volume and let $⟨r^2⟩$ be the mean squared radius of one electron. Use the symmetric magnetic vector potential $\mathbf{A}=\mathbf{B}\times \mathbf{r}/2$. Expanding the electron kinetic Hamiltonian $(\mathbf{p}+e\mathbf{A}{)}^2/(2m_e)$ gives the quadratic field termwhere isotropy gives $⟨x^2+y^2⟩=2⟨r^2⟩/3$. The term linear in orbital angular momentum vanishes in the spherical electronic ground state. ThusFor weak susceptibility, $B\simeq {\mu }_{0}H$, so the Langevin diamagnetic susceptibility isIf the supplied density is a mass density ${\rho }_4$, use $n_4={\rho }_4/(4u)$, with $u$ the atomic mass unit. The result is approximately temperature independent at fixed density. The factor two for the two electrons is already included; $⟨r^2⟩$ has not been defined as their summed radius.