The number density of electron is a semiconductor is `8xx10^(13) cm^(-3)` and number density of hole is `5xx10^(12) cm^(-3)` then,
(1) Which type of this semiconductor is?
(2) How much is resistivity of it? (Mobility of electron is 23000 `cm^(2)V^(-1)s^(-1)` and mobility of hole is 100 `cm^(2)V^(-1)s^(-1) and e=1.6xx10^(-19)C`)

Options

A. `p-29.45xx10^(-2)Omegam`
B. `n-3.396xx10^(-2)Omegam`
C. `p-3.396xx10^(-2)Omegam`
D. `n-2.945xx10^(-2)Omegam`

Answer

B

Explanation

`n-3.396xx10^(-2)Omegam`
`n_(e )=8xx10^(13)cm^(-3)=8xx10^(19)m^(-3)`
`n_(h)=5xx10^(12) cm^(-3)=5xx10^(18)m^(-3)`
`mu_(e )=23000 cm^(2)V^(-1)s^(1)=2.3m^(2)V^(-1)s^(-1)`
`mu_(h)=100cm^(2)V^(-1)s^(-1)=0.01m^(2)V^(-1)s^(-1)`
Here, the number of electrons is higher than the holes so semiconductor will be n-type semiconductor `(n_(e ) gt n_(h))`
`sigma=(1)/(rho)=e(mu_(e )mu_(e )+n_(h)mu_(h))=1.6xx10^(-19)`
`[8xx10^(19)xx2.3xx5xx10^(18)xx0.01]`
`=1.6xx10^(-19)[1.84xx10^(20)+5xx10^(16)]`
`(1)/(rho) ~~29.45"mho m"^(-1)`
`therefore rho=(1)/(29.45)=3.396xx10^(-2)`ohm m
`therefore rho=3.396xx10^(-2)Omegam`