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

According to the question,
`v_d=2.5 times 10^-4 m//s implies n=8 times 10^28//m^2`
We know that `J=nev_d or I=nev_dA`
where symbols have their usual meaning
`implies V/R=nev_d A`
or `V/((pL)/A)=nev_d A or V/(pL)=nev_D or p=V/(nev_dL)` `=5/(8 times 10^28 times 1.6 times 10^-19 times 2.5 times 10^-4 times 0.1) or p=1.6 times 10^-8 Omega m`