When 5 V potential difference is applied across a wire of length 0.1 m, the drift speed of electrons is `2.5 xx 10^(-4) ms ^(-1)`. If the electron density in the wire is ` 8 xx 10^(28) m^(-3)`, the resistivity of the material is close to :

Options

A. `1.6 xx 10^(-8) Omega m`
B. `1.6 xx 10^(-7) Omega m`
C. `1.6 xx 10^(-6) Omega m`
D. `1.6 xx 10^(-5) Omega m`

Answer

D

Explanation

5V potential difference is applied across the length of 0.1m. So the electric field intensity can be written as:
`E= V//l = 5//0.1 = 50V//m`
Other given entities are:
Drift velocity `v_(d) = 2.5 xx 10^(-4) m//s`
Electron density `n= 8 xx 10^(28) m^(-3)`
Drift velocity can be written as follows:
`v_(d) = (e Etau)/(m)`
Here `tau` is the relaxation time. We can write `tau` as follows:
`tau = (mv_(d))/(eE)`...(i)
Resistivity of the material can be written as follows:
`rho = (m)/(n e^(2) tau)`...(ii)
Substituting relaxation time from equation (i) into equation (ii) we get
`rho = (m)/(n e^(2) ((mv_(d))/(eE))) rArr rho = (E )/(n ev_(d))`
Substituting values in above found result we can calculate resistivity as follows:
`rho = (E )/(n ev_(d)) = (50)/((8 xx 10^(28)) (1.6 xx 10^(-19)) (2.5 xx 10^(-4)))`
`=1.6 xx 10^(-5)Omega m`
Hence option (d) is correct