Magnetic Properties of Solids Questions and Answers

Answer: C

The three behaviours correspond to \(\theta = 0, \theta \lt 0, \theta \gt 0\).

With \(\chi \propto 1/(T + \theta)\):

• A paramagnet obeys Curie’s law, \(\chi \propto 1/T\), so \(\theta = 0\).
• A ferromagnet above its Curie temperature obeys the Curie-Weiss law \(\chi \propto 1/(T - T_{C})\), which diverges at \(T_{C}\); this is \(\theta = -T_{C} \lt 0\).
• An antiferromagnet above its Néel temperature follows \(\chi \propto 1/(T + \theta)\) with \(\theta \gt 0\), so the susceptibility stays finite and the extrapolated intercept is at negative temperature.

The options with \(\theta \gt 0\) for the ferromagnet reverse the signs of the two ordered cases, and any option giving the paramagnet a non-zero \(\theta\) contradicts Curie’s law.

Answer: B

The combination is (P) - Paramagnet; (Q) - Diamagnet; (R) - Type-I Superconductor.

The slope of each line is the susceptibility \(\chi = M/H\). P has a small positive slope, the weak positive response of a paramagnet. Q has a small negative slope, the weak opposing response of a diamagnet. R has a slope of exactly -1, so \(M = -H\) and \(B = \mu_{0}(H + M) = 0\): perfect diamagnetism, the Meissner state of a Type-I superconductor.

S is steeper than -1, a susceptibility below -1, which no material shows in equilibrium. An antiferromagnet has a small positive susceptibility, so calling Q an antiferromagnet is wrong, and the options assigning S to an antiferromagnet or a Type-I superconductor, or calling P diamagnetic, fail as well.

Answer: A

The ratio is \(\frac{2\ln 2}{\ln(\frac{3}{2})}\).

For \(J = \frac{1}{2}\) there are two levels at \(\mp\mu B\). With \(x = \frac{\mu B}{k_{B}T}\), the fraction in the lower level is \(\frac{1}{1 + e^{-2x}}\).

\[ \begin{aligned} 80\%: e^{-2x_{1}} &= \tfrac{1}{4} \Rightarrow x_{1} = \ln 2 \\ 60\%: e^{-2x_{2}} &= \tfrac{2}{3} \Rightarrow x_{2} = \tfrac{1}{2}\ln\tfrac{3}{2} \end{aligned} \]

At the same temperature the field is proportional to \(x\), so \(\frac{B_{1}}{B_{2}} = \frac{2\ln 2}{\ln(3/2)}\), about 3.4.

\(\frac{\ln 2}{\ln(3/2)}\) drops the factor of 2 from \(x_{2}\), and the others come from wrong occupation fractions.

Answer: A

Above the Néel temperature the susceptibility has the form \( \chi = \frac{2c}{T + \theta} \), with the paper's constant written as 2c.

In the paramagnetic region an antiferromagnet obeys a Curie-Weiss law with a negative Weiss temperature, \( \chi = \frac{C}{T + \theta} \): the antiparallel exchange opposes alignment, so \( \chi \) stays finite at \( T = \theta \) and peaks at the Néel temperature.

\( \frac{C}{T - \theta} \) is the ferromagnetic Curie-Weiss law, which diverges at the Curie point. The forms \( 2c(T \pm \theta) \) increase with temperature, while a paramagnetic susceptibility falls as T rises.

Answer: C

J equals S when the shell is just half full.

By Hund's rules, electrons first occupy separate orbitals with parallel spins. In a half-filled shell every \( m_{l} \) value is used exactly once, so \( L = \sum m_{l} = 0 \) and \( J = |L \pm S| = S \). Half-filled ions such as Mn²⁺ (3d⁵, \( ^{6}S_{5/2} \)) show this.

For a shell less than half full the third rule gives \( J = |L - S| \), for more than half full \( J = L + S \), and for a completely full shell \( L = S = J = 0 \).

Answer: B

The ion has 3 unpaired 4f electrons.

In the term \(^{4}I_{9/2}\), the multiplicity 2S + 1 = 4 gives S = 3/2, which means three unpaired electrons each contributing 1/2. The letter I means L = 6, and J = L - S = 9/2, as Hund's third rule requires for a shell less than half full.

This is the ground term of \(Nd^{3+}\), which has the configuration \(4f^{3}\). Two unpaired electrons would give a triplet, four a quintet and five a sextet.