Ncert Exercises & Solutions

Explain quantitatively the order of magnitude difference between the diamagnetic susceptibility of $N_{2} (~ 5 \times 10^{- 9})$ (at STP) and $C u (~ 10^{- 5})$ .

Hint: The susceptibility of the materials depends on the density of the materials.

Step 1:

\begin{array}{l} We know that: \\ Density of nitrogen, ρ_{N_{2}} = \frac{28 g}{22 .4 L} = \frac{28 g}{22400 cc} \\ Also, density of copper, ρ_{Cu} = 8 g / cc \end{array}

Now, comparing both densities;

\frac{ρ_{N_{2}}}{ρ_{Cu}} = \frac{28}{22400} \times \frac{1}{8} = 1 .6 \times 10^{- 4}

Also given; \frac{χ_{N_{2}}}{χ_{Cu}} = \frac{5 \times 10^{- 9}}{10^{- 5}} = 5 \times 10^{- 4}

Step 2:

We know that,

χ = \frac{Magnetisation (M)}{Magnetic intensity (H)}

= \frac{Magnetic moment (M) / Volume (V)}{H}

= \frac{M}{HV} = \frac{M}{H (mass / density)} = \frac{Mρ}{Hm}

∴ χ \propto ρ (∵ \frac{M}{Hm} = constant)

Hence, \frac{χ_{N_{2}}}{χ_{Cu}} = \frac{ρ_{N_{2}}}{ρ_{Cu}} = 1 .6 \times 10^{- 4}

Thus, we can say that the magnitude difference or major difference between the diamagnetic susceptibility of

N_{2}

and Cu is due to their density.

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