The saturation magnetic induction of nickel is 0.65 weber/meter2. If the density of nickel is 8906 kg/m3 and atomic weight is 58.7, calculate the magnetic moment of the nickel atom in bohr magneton?
Answers
Given info : The saturation magnetic induction of nickel is 0.65 weber/meter2. If the density of nickel is 8906 kg/m3 and atomic weight is 58.7.
To find : The magnetic moment of the nickel atom on Bohr magneton is ..
solution : atomic weight of nickel, M = 58.7 g/mol
density of nickel, d = 8906 kg/m³
saturation magnetic induction of nickel, Ba = 0.65 Weber/m²
no of nickel atoms per m³, N = d × Avogadro's number/atomic weight
= 8906 × 6.023 × 10²³/58.7 × 10¯³
= 913.87 × 10²⁶ ≈ 914 × 10²⁶
now from formula,
saturation magnetisation, Ba = Nμ₀μ_m
where μ_m is magnetic moment
so, μ_m = Ba/Nμ₀
= 0.65/(914 × 10²⁶ × 4π × 10^-7)
= 5.66 × 10¯²⁴ Am²
9.29 × 10¯²⁴ Am² = 1 Bohr magneton
so, magnetic moment = 5.66/9.29 = 0.61 BM
Therefore the magnetic moment in Bohr magneton is 0.61