= Solution
{c}
Let $\mu_a=2.13\mu_N$ be the magnitude of the saturated nuclear moment of <helium-3>, with $\mu_N=e\hbar/(2m_p)$. A spin-one-half nucleus has the two Zeeman energies $\mp\mu_aB$. For independent nuclei the one-site <partition function> and magnetization density are
$$
z=2\cosh\left(\frac{\mu_aB}{k_BT}\right),\qquad M=n_3\mu_a\tanh\left(\frac{\mu_aB}{k_BT}\right).
$$
In the weak-field limit $\mu_aB\ll k_BT$, expand the hyperbolic tangent:
$$
\boxed{\chi_{3,\mathrm{nuc}}=\frac{\mu_0n_3(2.13\mu_N)^2}{k_BT}=\frac{\mu_0\rho_3(2.13\mu_N)^2}{3u\,k_BT}.}
$$
The nuclear moment's negative sign means that its moment points oppositely to its spin; it does not make the thermal <magnetic susceptibility> negative. The response is <paramagnetism> governed by <Curie's law>.
There is no extra factor $1/3$ here: the stated $2.13\mu_N$ is the moment of a fully polarized spin-one-half state. Equivalently, writing the moment operator as $g_N\mu_N\mathbf I$ requires $g_N=-4.26$, and $g_N^2\mu_N^2I(I+1)/3=(2.13\mu_N)^2$. The formula presumes independent localized nuclear moments; exchange correlations invalidate it close to the much lower ordering scale discussed in part (e).
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