= Solution
{c}
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 $\langle r^2\rangle$ 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 term
$$
\Delta E_{\mathrm{atom}}=\sum_{a=1}^2\frac{e^2B^2}{8m_e}\langle x_a^2+y_a^2\rangle=\frac{2e^2B^2}{12m_e}\langle r^2\rangle,
$$
where isotropy gives $\langle x^2+y^2\rangle=2\langle r^2\rangle/3$. The term linear in orbital angular momentum vanishes in the spherical electronic ground state. Thus
$$
M=-n_4\frac{\partial\Delta E_{\mathrm{atom}}}{\partial B}=-\frac{n_4e^2\langle r^2\rangle}{3m_e}B.
$$
For weak susceptibility, $B\simeq\mu_0H$, so the <Langevin diamagnetic susceptibility> is
$$
\boxed{\chi_4=-\frac{\mu_0n_4e^2\langle r^2\rangle}{3m_e}.}
$$
If 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; $\langle r^2\rangle$ has not been defined as their summed radius.
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