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The number of formula units per volume follows from the mass density:Each neutral formula unit contains $31+33=64$ electrons, so ${\rho }_e=64n_f=1.5726\times 10^{30}{\mathrm{m}}^{-3}$. Let $R^2$ be the electron-weighted mean-square orbital radius, averaged over all these electrons. The supplied Larmor-Langevin susceptibility then gives the dimensionless SI volume magnetic susceptibilitywhere $a_0=0.0529177\mathrm{nm}$ is the Bohr radius. An explicit rough atomic-scale assumption $R\simeq a_0$ gives ${\chi }_0\simeq -2.6\times 10^{-5}$.

The quoted bulk data do not determine $R$: a nuclear charge does not specify an orbital radius in a screened many-electron chemical atom. Consequently the numerical estimate just made has atomic-model uncertainty; it is not a deduction with guaranteed 30 percent accuracy. For example, the alternative assumed $R=2a_0$ gives four times the magnitude while leaving every supplied bulk number unchanged. Atomic wavefunctions, an explicit screened-orbital model, or measured atomic magnetic susceptibilities of materials exhibiting diamagnetism would be needed to support that accuracy. Keeping $R$ explicit makes the later comparison independent of an unadvertised radius assumption. The bare electron mass belongs in this atomic formula; the conduction band effective mass is relevant instead to the large donor envelope in part (c).

Use the stipulated noninteracting impurity band approximation, with one donated electron per donor and two available spin states per donor orbital. The energy band is half filled. Approximating its spin-summed density of states per volume near the Fermi energy by $D_v(E_F)\simeq 2n/W$ gives the zero-temperature Pauli paramagnetismWith $\alpha =1$,A factor of order one depends on the density of states shape; using the conduction band $g$-factor rather than silently setting $g=2$ matters here.

To locate the comparison with the magnitude of the undoped Larmor-Langevin susceptibility, put $x=n^{-1/3}/a_b>1$. Numerically,For the stated atomic-scale estimate $R=a_0$ in part (b), equality with $|{\chi }_0|=2.599\times 10^{-5}$ requiresThus the dilute-branch crossover iswell below $a_b^{-3}=9.45\times 10^{23}{\mathrm{m}}^{-3}$. Within this model the positive spin response exceeds the negative background magnitude for lower dopant densities on this dilute branch: the exponential narrowing of $W$ wins over the loss of donors. Indeed $d\ln (e^x/x^3)/dx=1-3/x>0$ for $x>3$. This is not a claim of ferromagnetism.

The degenerate Pauli paramagnetism calculation requires $k_{B}T\ll W$, with an unimportant order-one filling factor. At the estimated crossing,The weak-field condition also requires $|g|{\mu }_{B}B\ll W$. For a different assumed atomic $R$, use $|{\chi }_0|=2.599\times 10^{-5}(R/a_0{)}^2$ in the displayed equation for $x$. An order-of-magnitude background $|{\chi }_0|\sim 10^{-5}$ gives the same density order and a crossover scale of order $10^{-5}\mathrm{K}$. At higher temperature than this extremely narrow bandwidth permits, the $n/W$ enhancement cannot be used; the independent-donor Curie law is the appropriate thermal scale if interactions remain negligible. At very small dopant density, disorder, localization and Coulomb interactions can also invalidate the assumed noninteracting impurity band. The formal divergence as $n\to 0$ is therefore a limit of the stated zero-temperature band model, not a prediction for isolated donors at a fixed nonzero temperature.