In semiconductor the concentrations of electron and holes are `8xx10^(18)//m^(3)` and `5xx10^(18)//m` respectively. If the mobilities of electrons and hole are `2.3 m^(2)//`volt-sec and `0.01m^(2)//` volt-sec respectively, then semicondutor is

Options

A. `N`-type and its resistivity is `0.34` ohm-meter
B. `p`-type and its resistivity is `0.034` ohm -meter
C. `N`-type and is resistivity is `0.034` ohm-meter
D. `P`-type and its resistivity is `3.40` ohm-meter

Answer

A

Explanation

`n_(e)=8xx10^(18)//m^(3),n_(h)=5xx10^(18)//m^(3)`
`mu_(e) =2.3 (m^(2))/(vol t-sec), mu_(h)=0.01 (m^(2))/(vol t-sec)`
`:' n_(e)gtn_(h)` so semiconductor is `N`-type
Also conductivity `sigma=1/(Resistivity (rho))=e(n_(e)mu_(e)+n_(h)mu_(h))`
`implies 1/(rho)=1.6xx10^(-19)[8xx10^(18)xx2.3+5xx10^(18)xx0.01]`
`rho=0.34 Omega-m`.