When 5V 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 times 10^-8 Omega m`
B. `1.6 times 10^-7 Omega m`
C. `1.6 times 10^-6 Omega m`
D. `1.6 times 10^-5 Omega m`

Explanation

`V=IR=Jp I/A implies p=(VA)/(II)=(VA)/(In eAv_D)=V/(l times n times e times v_d)`
`implies p=5/(0.1 times 2.5 times 10^-19 times 1.6 times 10^-19 times 8 times 10^28)=1.6 times 10^-5 Omega -m`