Discussion
A paramagnetic material containing paramagnetic ions with total angular momentum \(J = \frac{1}{2}\) is kept at absolute temperature T. The ratio of the magnetic field required for 80% of the ions to be in the lowest energy state to that required for having 60% of the ions to be in the lowest energy state at the same temperature is
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.
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