InterviewSolution
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InterviewSolution
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Explain quantitatively the order of magnitude difference between the diamagnetic susceptibility of `N_2(-5xx10^-9)` (at STP) and `Cu(-10^-5)`. |
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Answer» Density of nitrogen, `rho_(N_2)=(28g)/(22.4litre)=(28g)/(22400c c)` We know, density of copper, `rho_(cu)=8g//c c` `:. (rho_(N_2))/(rho_(Cu))=(28)/(22400)xx1/8=1*6xx10^-4`. Given, `(chi_(N_2))/(chi_(cu))=(5xx10^-9)/(10^-5)=5xx10^-4` …(i) We know that `chi= ("intensity of magnetisation"(I))/("magnetising force" (H))=I/H=("magnetic moment"(M)//volume(V))/(H)` `=(M)/(HV)=(M)/(H["mass"(m)//"density"(rho)])=(Mrho)/(Hm)` `:. chiproprho` for the given value of `M//Hm`. Hence, `(chi_(N_2))/(chi_(Cu))=(rho_(N_2))/(rho_(Cu))=1*6xx10^-4` ...(ii) From (i) and (ii), we note that the major difference in susceptibility of `N_2` and `Cu` is due to their density. |
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