Question 6 of 7: Diamagnetism vs. Paramagnetism, and the Langevin Susceptibility

Question 6: Diamagnetism vs. Paramagnetism, and the Langevin Susceptibility (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.

Find. (a) the physical distinction between diamagnetism and paramagnetism; (b) the magnetic susceptibility $\chi$ of the described diamagnetic material.

Approach. (a) contrast the origin of the induced vs. permanent magnetic response. (b) recognize the atom as helium (2p+2n+2e, atomic weight 4 amu; the given density and radius match gaseous He at STP), compute the number density $N/V$, then apply the Langevin diamagnetic susceptibility formula, eq. (17).

1. Part (a) — diamagnetism vs. paramagnetism. Diamagnetism is present in every material: an applied field induces a Larmor precession in each atom's closed (filled) electron orbits, and by Lenz's law this induced circulating current opposes the applied field, giving a small, negative, essentially temperature-independent susceptibility ($\chi<0$, $|\chi|\sim10^{-5}$ to $10^{-10}$). Paramagnetism occurs only in materials whose atoms carry a net permanent magnetic moment (unpaired electrons, e.g. partially-filled orbitals); an applied field partially aligns these pre-existing moments against thermal randomization, giving a larger, positive susceptibility that falls with temperature (Curie's law, $\chi\propto1/T$). Where both mechanisms coexist, paramagnetism (when present) normally dominates and masks the always-present diamagnetic background.
2. Part (b) — identify the material and its number density. With 2 protons + 2 neutrons per nucleus (atomic weight $\approx4\ \text{amu}$) and 2 electrons, this is helium; the stated density, $1.785\times10^{-4}\ \text{g/cm}^3$, is indeed the density of He gas at STP — an internal consistency check on the given data. Number density (SI): $$N=\frac{\rho N_A}{M}=\frac{(0.1785\ \text{kg/m}^3)(6.02217\times10^{23}\ \text{mol}^{-1})}{4\times10^{-3}\ \text{kg/mol}}$$ $$N\approx2.687\times10^{25}\ \text{atoms/m}^3$$
3. Part (b), cont'd — Langevin susceptibility, eq. (17). With $Z=2$ electrons/atom, $\langle r^2\rangle\approx r^2=(3.1\times10^{-11}\ \text{m})^2=9.61\times10^{-22}\ \text{m}^2$, $\mu_0=4\pi\times10^{-7}\ \text{N/A}^2$, $m=9.10956\times10^{-31}\ \text{kg}$ (electron rest mass, per the constants page): $$\chi=-\frac{\mu_0Ne^2Z\langle r^2\rangle}{6m}=-\frac{(4\pi\times10^{-7})(2.687\times10^{25})(1.60219\times10^{-19})^2(2)(9.61\times10^{-22})}{6(9.10956\times10^{-31})}$$ $\boxed{\chi\approx-3.05\times10^{-10}}$ (dimensionless, SI volume susceptibility) — negative, as required for a diamagnetic response, and of the correct order of magnitude for a dilute gas (compressed condensed-phase diamagnets are typically $10^3$–$10^5$ times larger since $N$ scales up accordingly).