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If the resistivity of copper is `1.7 xx 10^(-6) Omega cm`, then the mobility of electrons in copper, if each atom of copper contributes one free electron for conduction, is [The amomic weight of copper is `63.54` and its density is `8.96 g//c c]` :A. 23.36 `cm^(2)`/VsB. 503.03 `cm^(2)`/VsC. 43.25 `cm^(2)`/VsD. 88 `cm^(2)`/Vs |
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Answer» Correct Answer - C Mobility of electron, `mu=simga/(ne)` …………..(i) Resistivity, `rho=1/sigma`………..(ii) From Eqs. (i) and (ii), we get `mu=1/(nep)`………….(iii) n=number of free electrons per unit volume `n=(N_(0) xx d)/("Atomic weight")` = `(6.023 xx 10^(23) xx 8.96)/(63.54)` `=8.5 xx 10^(22)`...............(iv) From Eqs. (iii) and (iv), we get `mu = (1)/(8.5 xx 10^(22) xx 1.6 xx 10^(-19) xx 1.7 xx 10^(-6))=43.25 cm^(2)//Vs` |
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