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If the resistivity of copper is 1.7 xx 1...

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)`/Vs

B

503.03 `cm^(2)`/Vs

C

43.25 `cm^(2)`/Vs

D

88 `cm^(2)`/Vs

Text Solution

Verified by Experts

The correct Answer is:
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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