Consider an atomic gas with number density $ n = 10^{20} \, \text{m}^{-3} $, in the ground state at 300 K. The valence electronic configuration of atoms is $ f^7 $. The paramagnetic susceptibility of the gas $ \chi = m \times 10^{-11} $. The value of $ m $ (rounded off to two decimal places) is $\underline{\hspace{2cm}}$. (Given: Magnetic permeability of free space $ \mu_0 = 4\pi \times 10^{-7} \, \text{H m}^{-1} $, Bohr magneton $ \mu_B = 9.274 \times 10^{-24} \, \text{A m}^2 $, Boltzmann constant $ k_B = 1.3807 \times 10^{-23} \, \text{K}^{-1} $)

Question

Consider an atomic gas with number density $ n = 10^{20} \, \text{m}^{-3} $, in the ground state at 300 K. The valence electronic configuration of atoms is $ f^7 $. The paramagnetic susceptibility of the gas $ \chi = m \times 10^{-11} $. The value of $ m $ (rounded off to two decimal places) is $\underline{\hspace{2cm}}$.   (Given: Magnetic permeability of free space $ \mu_0 = 4\pi \times 10^{-7} \, \text{H m}^{-1} $, Bohr magneton $ \mu_B = 9.274 \times 10^{-24} \, \text{A m}^2 $, Boltzmann constant $ k_B = 1.3807 \times 10^{-23} \, \text{K}^{-1} $)

cdquestions Admin · Accepted Answer

The paramagnetic susceptibility is related to the magnetic moment by: $$ \chi = \frac{\mu^2 n}{k_B T}, $$ where $ T = 300 \, \text{K} $ and $ \mu = m \times \mu_B $. Using the given values and solving for $ m $, we obtain $ m \approx 5.4 $. Thus, the value of $ m $ is $ 5.40 $.

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Correct Answer: 5.4

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