Question 6 of 7: Paramagnetism, Diamagnetism, and the Molar Susceptibility of an Inert Gas

Question 6: Paramagnetism, Diamagnetism, and the Molar Susceptibility of an Inert Gas (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Part (a) — paramagnetism. Paramagnetism arises in atoms/ions that possess a net permanent magnetic moment — from unpaired electron spins and/or unquenched orbital angular momentum (e.g. partially filled orbital sets, as in transition-metal and rare-earth ions, or a lone unpaired electron). With no applied field these moments point randomly (thermal disorder), giving zero net magnetization; an applied field $B$ partially aligns them, producing a small POSITIVE susceptibility that follows the Curie law $\chi\propto1/T$ — it strengthens the applied field and gets weaker as temperature rises because thermal agitation increasingly randomizes the alignment.

Part (b) — diamagnetism. Diamagnetism is the induced response of the orbiting (filled-orbital) electrons themselves: by Lenz's law/Larmor precession, an applied field induces a tiny extra circulating current in every electron orbit that opposes the change in flux through it, producing a magnetic moment directed opposite to the applied field. This gives a small NEGATIVE, essentially temperature-independent susceptibility, and — unlike paramagnetism — it is present in every atom (filled-orbital or not); it is simply masked by the much larger paramagnetic (or ferromagnetic) response whenever unpaired moments are present.

Given (c). $\rho=0.214\ \text{g/cm}^3$, $\langle r\rangle\approx3.8\times10^{-9}\ \text{cm}=3.8\times10^{-11}\ \text{m}$ (taken as $\sqrt{\langle r^2\rangle}$), $Z=2$ electrons (2 protons, 2 neutrons $\Rightarrow$ atomic mass $\approx4\ \text{amu}$, i.e. this is helium). Paper's Langevin-diamagnetism formula, eq. (17): $\chi=-\mu_0NZe^2\langle r^2\rangle/(6m)$.

Find (c). the molar susceptibility $\chi$ in cm³/mol, and whether the gas is paramagnetic or diamagnetic.

Approach. Read eq. (17) with $N$ as the actual atomic number density (SI, m$^{-3}$), giving a dimensionless volume susceptibility; convert to a per-mole quantity by multiplying by the molar volume $V_m=M/\rho$, then apply the standard SI$\to$CGS volume-susceptibility conversion (divide by $4\pi$) to land in the conventional cm³/mol unit the question asks for.

1. Step 1 — number density and volume susceptibility. Atomic mass $M=4\ \text{amu}=4(1.66053\times10^{-27}\ \text{kg})=6.6421\times10^{-27}\ \text{kg}$ per atom. Number density: $$n=\frac{\rho}{M}=\frac{214\ \text{kg/m}^3}{6.6421\times10^{-27}\ \text{kg}}=3.222\times10^{28}\ \text{m}^{-3}$$ Substituting into eq. (17) (SI, $N\to n$) with $\mu_0=4\pi\times10^{-7}\ \text{N/A}^2$, $e=1.60219\times10^{-19}\ \text{C}$, $\langle r^2\rangle=(3.8\times10^{-11}\ \text{m})^2=1.444\times10^{-21}\ \text{m}^2$, $m=9.10956\times10^{-31}\ \text{kg}$: $$\chi_v=-\mu_0nZe^2\langle r^2\rangle/(6m)=-5.49\times10^{-7}\ \text{(dimensionless, SI)}$$
2. Step 2 — convert to molar susceptibility, cm³/mol. Molar volume $V_m=M\,N_A/\rho=(6.6421\times10^{-27})(6.02217\times10^{23})/214=1.869\times10^{-5}\ \text{m}^3/\text{mol}=18.69\ \text{cm}^3/\text{mol}$. The SI molar susceptibility is $\chi_v\times V_m=-1.027\times10^{-11}\ \text{m}^3/\text{mol}$; converting to the conventional CGS cm³/mol unit (divide the SI volume susceptibility by $4\pi$, then express the volume in cm³): $$\chi_{\text{molar}}=\frac{\chi_v\,V_m}{4\pi}\times10^6\ \text{cm}^3/\text{m}^3$$ $$\boxed{\chi_{\text{molar}}\approx-8.17\times10^{-7}\ \text{cm}^3/\text{mol}}$$ (the density cancels between $n\propto\rho$ and $V_m\propto1/\rho$, so this is equivalently just $-\mu_0N_AZe^2\langle r^2\rangle/(6m\cdot4\pi)$ in cm³/mol — a genuinely density-independent, per-atom property, as a molar susceptibility should be). This is the same order of magnitude as the accepted experimental molar susceptibility of helium gas, $\chi_{\text{molar}}\approx-1.9\times10^{-6}\ \text{cm}^3/\text{mol}$, which is a reasonable check given the crude "average radius" model used here for $\langle r^2\rangle$ in place of the true quantum-mechanical expectation value.
3. Step 3 — sign/classification. $\chi<0$ throughout, i.e. the induced moment opposes the applied field. $$\boxed{\text{The gas is DIAMAGNETIC}}$$ consistent with helium's closed $1s^2$ configuration: no unpaired electrons are available to give a competing (positive) paramagnetic contribution.