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:
Show Hint
- \( \frac{2 \ln 2}{\ln \left(\frac{3}{2}\right)} \)
- \( \frac{\ln 2}{\ln \left(\frac{3}{2}\right)} \)
- \( \frac{3 \ln 2}{\ln \left(\frac{3}{2}\right)} \)
- \( \frac{\ln 3}{\ln \left(\frac{3}{2}\right)} \)
The Correct Option is A
Solution and Explanation
In the case of paramagnetic materials, the population of ions in the energy states is governed by the Boltzmann distribution.
The probability of an ion being in the lowest energy state is given by: \[ P = \frac{1}{1 + e^{\frac{E}{kT}}} \] where \( E \) is the energy difference between the states, \( k \) is Boltzmann's constant, and \( T \) is the temperature. For paramagnetic ions with angular momentum \( J = \frac{1}{2} \), the energy difference \( E \) between the states is proportional to the magnetic field \( B \), i.e., \( E \propto B \). So, the ratio of magnetic fields required for different probabilities of being in the lowest energy state follows the relation: \[ \frac{B_1}{B_2} = \frac{\ln P_2}{\ln P_1} \] For \( P_1 = 0.8 \) and \( P_2 = 0.6 \), we calculate the ratio as: \[ \frac{B_1}{B_2} = \frac{\ln 0.6}{\ln 0.8} = \frac{2 \ln 2}{\ln \left(\frac{3}{2}\right)} \] Thus, the correct answer is \( \frac{2 \ln 2}{\ln \left( \frac{3}{2} \right)} \).
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