Question 6 of 7: Magnetism in Crystal Lattices — Diamagnetism, Molar Susceptibility, Curie Constant

Question 6: Magnetism in Crystal Lattices — Diamagnetism, Molar Susceptibility, Curie Constant (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.

Given. Molar susceptibilities of five inert gases (all negative, part b). Hypothetical inert gas for part (c): atomic mass $4.002\ \text{g/mol}$, density $0.214\ \text{g/cm}^3$, $Z=2$ electrons, average atomic radius $r=1.7\times10^{-9}\ \text{cm}$. Figure P6: $\chi^{-1}$ (in $10^4\ \text{mole/cm}^3$) vs. $T$ (K), measured points with a fitted straight line.

Find. (a) definition of electron spin; (b) para- vs. diamagnetic classification and physical source; (c) molar susceptibility $\chi_{molar}$; (d)(i) classification from the graph and physical explanation; (d)(ii) Curie constant $C$.

Approach. (b)/(c) apply the Langevin diamagnetism formula, eq. (17); (d) read the Curie-law slope $\chi^{-1}=T/C$ directly off Figure P6.

1. Part (a) — electron spin. Electron spin is the electron's intrinsic angular momentum, of fixed magnitude $\hbar\sqrt{s(s+1)}$ with $s=\tfrac12$, existing in addition to any orbital angular momentum and with no classical analogue (it is not literally the electron spinning about an axis). It is quantized along any measurement axis to $\pm\hbar/2$, and it carries an associated intrinsic magnetic moment of magnitude one Bohr magneton, $\mu_B=e\hbar/2mc$.
2. Part (b) — inert gases: diamagnetic. All five listed susceptibilities are negative, so all five inert gases are diamagnetic. Each atom has a filled, closed-orbital electron configuration with zero net orbital and spin angular momentum, so there is no permanent atomic magnetic moment to align with an applied field. Instead, by Lenz's law, the applied field induces a small circulating current in each closed orbital that opposes the applied flux (Larmor precession of the whole electron cloud), producing a small magnetization opposed to the field — hence $\chi<0$. The magnitude grows monotonically with atomic number (He $\to$ Xe) because $\chi\propto Z\langle r^2\rangle$ (eq. 17): heavier noble gases have more electrons in progressively larger orbitals.
3. Part (c) — molar susceptibility from atomic data. Eq. (17) in CGS form, evaluated with $N=N_A$ (Avogadro's number) to give a molar susceptibility directly: $$\chi_{molar}=-\frac{N_AZe^2}{6mc^2}\langle r^2\rangle$$ With $Z=2$, $e=4.803\times10^{-10}\ \text{esu}$, $m=9.11\times10^{-28}\ \text{g}$, $c=2.998\times10^{10}\ \text{cm/s}$, $r=1.7\times10^{-9}\ \text{cm}$: $$\chi_{molar}=-\frac{(6.022\times10^{23})(2)(4.803\times10^{-10})^2}{6(9.11\times10^{-28})(2.998\times10^{10})^2}(1.7\times10^{-9})^2\ \text{cm}^3/\text{mol}$$ $$\boxed{\chi_{molar}\approx-1.6\times10^{-7}\ \text{cm}^3/\text{mol}}$$ Note the atomic mass and density cancel out of this formula: writing the atomic number density as $n=\rho N_A/M_{atomic}$ and multiplying the volume susceptibility $-n Ze^2\langle r^2\rangle/6mc^2$ by the molar volume $M_{atomic}/\rho$ to convert it to a molar quantity reproduces exactly the $N_A$-only expression above — so density and atomic mass are given data that are not actually needed once the target quantity is per mole. The result is negative, as it must be, but about an order of magnitude smaller than the measured He value in part (b) ($-1.9\times10^{-6}$). The difference comes entirely from the given radius: $\chi\propto\langle r^2\rangle$, and $1.7\times10^{-9}$ cm (0.17 Å) is much smaller than the real rms radius of the He electron cloud. Reproducing $-1.9\times10^{-6}$ would need $r\approx5.8\times10^{-9}$ cm (0.58 Å).
4. Part (d)(i) — classification and $T$-dependence. Figure P6 plots $\chi^{-1}$ increasing linearly with $T$, extrapolating to (very nearly) the origin — i.e. $\chi=C/T$, the Curie law, which describes a paramagnetic crystal (permanent atomic magnetic moments, unlike the induced-only diamagnetism of Part (b)). At high temperature, thermal agitation randomizes the orientation of the individual permanent moments faster than the applied field can align them, so the net magnetization — and hence $\chi$ — falls as $T$ rises; conversely, as $T\to0$ thermal randomization vanishes and $\chi$ diverges (in the ideal Curie-law limit), consistent with $\chi^{-1}\to0$ at the graph's origin.
5. Part (d)(ii) — Curie constant. The Curie law $\chi=C/T\Rightarrow\chi^{-1}=T/C$, so $1/C$ is the slope of the plotted line. Taking two well-separated points on the printed line, $(40\ \text{K},\,0.46\times10^4)$ and $(297\ \text{K},\,4.12\times10^4\ \text{mole/cm}^3)$: $$\frac1C=\text{slope}=\frac{(4.12-0.46)\times10^4\ \text{mole/cm}^3}{(297-40)\ \text{K}}\approx142\ \text{mole}\,\text{cm}^{-3}\text{K}^{-1}$$ $$\boxed{C\approx7.0\times10^{-3}\ \text{cm}^3\,\text{K}/\text{mole}}$$ A least-squares fit to the whole digitized line gives the same slope. Forcing the line through the origin, using the end point alone ($297/4.12\times10^{-4}$), gives $7.2\times10^{-3}$. A graph reading is good to a few percent, so quote $C\approx7\times10^{-3}\ \text{cm}^3\,\text{K/mole}$.