Mobility of electrons in a semiconductor is defined as the ratio of their drift velocity to the applied electric field. If, for an n-type semiconuctor, the density of electrons is `10^(19)m^(-3)` and their mobility is `1.6m^(2)//(V.s)` then the resistivity of te semiconductor (since it is an n-type semiconductor contribution of holes is ignored) is close to :

Explanation

As we know current density
`j=sigma E=n e v_(d)`
`sigma n e (v_(d))/E=n e mu`
`1/(sigma) = p 1/(n_(e)e mu_(e))=` Resistivity
`=1/(10^(19)xx1.6xx10^(19)-19xx1.6)`
or `P=0.4Omegam`