Question
Explain quantitatively the order of magnitude difference between the diamagnetic susceptibility of $N_2 (~5 \times 10^{-9})$ (at STP) and Cu $(~10^{-5}).$
✓
Answer
Key concept:
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., $\text{XM}=\frac{\text{I}}{\text{H}}$.
According to the problem, we have
Density of nitrogen $\rho_{\text{N}_2}=\frac{28\text{g}}{22.4\text{L}}=\frac{28\text{g}}{22400\text{cc}}$
Also, density of copper $\rho_{\text{Cu}}=\frac{8\text{g}}{22.4\text{L}}=\frac{8\text{g}}{22400\text{cc}}$
So, ration of both densities
$\frac{\rho_{\text{N}_2}}{\rho_\text{Cu}}=\frac{28}{22400}\times\frac{1}{8}=16\times10^{-4}$
Also given $\frac{\chi_{\text{N}_2}}{\chi_\text{Cu}}=\frac{5\times10^{-9}}{10^{-5}}=5\times10^{-4}$
we know that, $\chi=\frac{\text{Magentisation (M)}}{\text{Magnetic intensity (H)}}$
$=\frac{\text{Magnetic moment (M)/Volime (V)}}{\text{H}}$
$=\frac{\text{M}}{\text{HV}}=\frac{\text{M}}{\text{H}(\text{mass/density})}=\frac{\text{M}\rho}{\text{Hm}}$
$\chi\propto\rho\ \ \Big(\because\ \frac{\text{M}}{\text{Hm}}=\text{constant}\Big)$
Hence, $\frac{\chi_{\text{N}_2}}{\chi_\text{Cu}}=\frac{\rho_{\text{N}_2}}{\rho_\text{Cu}}=1.6\times10^{-4}$
Therefore, we can say that magnitude difference or major difference between the diamagnetic susceptibility of $N_2$ and Cu is $1.6 \times 10^{-4}.$
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., $\text{XM}=\frac{\text{I}}{\text{H}}$.
According to the problem, we have
Density of nitrogen $\rho_{\text{N}_2}=\frac{28\text{g}}{22.4\text{L}}=\frac{28\text{g}}{22400\text{cc}}$
Also, density of copper $\rho_{\text{Cu}}=\frac{8\text{g}}{22.4\text{L}}=\frac{8\text{g}}{22400\text{cc}}$
So, ration of both densities
$\frac{\rho_{\text{N}_2}}{\rho_\text{Cu}}=\frac{28}{22400}\times\frac{1}{8}=16\times10^{-4}$
Also given $\frac{\chi_{\text{N}_2}}{\chi_\text{Cu}}=\frac{5\times10^{-9}}{10^{-5}}=5\times10^{-4}$
we know that, $\chi=\frac{\text{Magentisation (M)}}{\text{Magnetic intensity (H)}}$
$=\frac{\text{Magnetic moment (M)/Volime (V)}}{\text{H}}$
$=\frac{\text{M}}{\text{HV}}=\frac{\text{M}}{\text{H}(\text{mass/density})}=\frac{\text{M}\rho}{\text{Hm}}$
$\chi\propto\rho\ \ \Big(\because\ \frac{\text{M}}{\text{Hm}}=\text{constant}\Big)$
Hence, $\frac{\chi_{\text{N}_2}}{\chi_\text{Cu}}=\frac{\rho_{\text{N}_2}}{\rho_\text{Cu}}=1.6\times10^{-4}$
Therefore, we can say that magnitude difference or major difference between the diamagnetic susceptibility of $N_2$ and Cu is $1.6 \times 10^{-4}.$
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