Consider a helium atom in its ground state (1s). The mean radius of the atom may be approximated by bohr radius 0.529å. The density of helium is 0.178 kg/m3. Calculate the diamagnetic susceptibility of helium atom using classical langevin formula.
Answers
The diamagnetic susceptibility of helium atom is -5.5 * 10^-5 m^3/mol
Given:
Mean radius = 0.529 A°
Density of helium = 0.178 kg/m3
To find:
Diamagnetic susceptibility of helium atom using classical langevin formula.
Solution:
The classical Langevin formula for diamagnetic susceptibility is given by:
χ_d = -(n/V) * (μ_B^2/3k_B) * (Z/M)
where n is the number density of atoms, V is the volume, μ_B is the Bohr magneton, k_B is the Boltzmann constant, Z is the atomic number, and M is the atomic mass.
To calculate the diamagnetic susceptibility of a helium atom using this formula, we need to know the number density of atoms, which can be calculated from the density of helium and the atomic mass of helium.
The number density of helium atoms = 0.178 kg/m3 / (4.003 g/mol) = 0.04465 mol/m3
The atomic number of helium is 2, the atomic mass of helium is 4.003 g/mol and the Bohr magneton is 9.27*10^-24 J/T
By inserting the known values into the Langevin formula, we get
χ_d = -(0.04465 mol/m3) * (9.2710^-24 J/T)^2 / (3 * 1.3810^-23 J/K) * (2/4.003 g/mol) = -5.5 * 10^-5 m^3/mol
Therefore, the diamagnetic susceptibility of helium atom is -5.5 * 10^-5 m^3/mol.
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