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Question
Explain quantitatively the order of magnitude difference between the diamagnetic susceptibility of N2 (~5 × 10–9) (at STP) and Cu (~10–5).
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Solution
Magnetic susceptibility: It is the property of the substance which shows how easily a substance can be magnetised. It can also be defined as the ratio of intensity of magnetisation (I) in a substance to the magnetic intensity (H) applied to the substance, i.e., XM = I/H.
According to the problem, we have
Density of nitrogen `ρ_(N_2) = (28 g)/(22.4 L) = (28 g)/(22400 "cc")`
Also. density of copper `ρ_(C_u) = (8 g)/(22.4 L) = (8 g)/(22400 "cc")`
So, ratio of both densities
`(ρ_(N_2))/(ρ_(Cu)) = 28/22400 xx 1/8 = 16 xx 10^-4`
Also given `(chi_(N_2))/(chi_(Cu)) = (5 xx 10^-9)/10^-5 = 5 xx 10^-4`
`(chi_m)_("Fero") ≈ 10^3`
and `(chi_m)_("Para") ≈ 10^-5`
⇒ `((chi_m)_("Fero"))/((chi_m)_("Para")) = 10^3/10^-5 = 10^8`
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