IGNOU MPH 12 2026 - For Chromium () vapour at with a number density of atoms&nb

Question

For Chromium ( $J = 3, S = 3, Z = 24$ ) vapour at $1800\ \text{K}$ with a number density of atoms $\frac{N}{V} = 5.0 \times 10^{25}\ \text{m}^{-3}$ calculate:
i) The Larmor diamagnetic susceptibility assuming the atomic radius to be 1.2 A.

ii) The Curie paramagnetic susceptibility.

19 Jan 2026

Answer :

Word Count : 626

We are asked to calculate Larmor diamagnetic susceptibility and Curie paramagnetic susceptibility for chromium vapor. Let’s solve it step by step manually. --- ### Given: * Chromium: (J = 3, S = 3, Z = 24) * Temperature: (T = 1800\ \text{K}) * Number density: ( \frac{N}{V} = 5.0 \times 10^{25}\ \text{m}^{-3} ) * Atomic radius: (r = 1.2\ \text{Å} = 1.2 \times 10^{-10}\ \text{m}) * Constants: * Permeability of free space: ( \mu_0 = 4 \pi \times 10^{-7}\ \text{H/m} ) * Bohr magneton: ( \mu_B = 9.274 \times 10^{-24}\ \text{A·m}^2 ) * Boltzmann constant: ( k = 1.38 \times 10^{-23}\ \text{J/K} ) --- ## (i) Larmor Diamagnetic Susceptibility Larmor diamagnetic susceptibility per unit volume is: [ \chi_L = -\frac{\mu_0 N e^2 r^2}{6 m_e} ] where: * (N) = number density * (e = 1.602 \times 10^{-19}\ \text{C}) * (r = 1.2 \times 10^{-10}\ \text{m}) * (m_e = 9.109 \times 10^{-31}\ \text{kg}) Step 1: Compute (e^2 r^2) [ e^2 = (1.602 \times 10^{-19})^2 = 2.566 \times 10^{-38} ] [ r^2 = (1.2 \times 10^{-10})^2 = 1.44 \times 10^{-20} ] [ e^2 r^2 = 2.566 \times 10^{-38} \times 1.44 \times 10^{-20} \approx 3.69 \times 10^{-58}

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