Question

Explain quantitatively the order of magnitude difference between the diamagnetic susceptibility of N₂ (~5 × 10^–9) (at STP) and Cu (~10^–5).

Answer 1

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., X_M = 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`