A semiconductor having electron and linear mobilities `mu_(n)` and `mu_(p)` respectively.
If its intrinsic carrier density is `n_(i)`, then what will be the value of hole concentration `P` for which the conductivity will be maximum at a given temperature?

Options

A. `n_(i)sqrt(mu_(n)/mu_(p))`
B. `n_(h)sqrt(mu_(n)/mu_(p))`
C. `n_(i)sqrt(mu_(p)/mu_(n))`
D. `n_(h)sqrt(mu_(p)/mu_(n))`

Answer

A

Explanation

Total conductivity of semiconductor is
`sigma=n_(e) e mu_(n)+n_(p) e mu_(p)=e(n_(e) mu_(n)+n_(p)mu_(p))`
We know that for an intrinsic semiconductor
`n_(e)n_(p)=n_(i)^(2) or n_(e)=n_(i)^(2)//n_(p)`
Hence, `sigma=e[n_(i)^(2)/n_(p)mu_(n)+n_(p)mu_(p)] ...(i)`
If `sigma` is minimum then `(dsigma)/(dn_(p))=0`
Differentiating (i) w.r.t. n_(p), we get
`(dsigma)/(dn_(p))=e[-n_(i)^(2)/n_(p)^(2)mu_(n)+mu_(p)]=0`
or `mu_(p)=n_(i)^(2)/n_(p)^(2)mu_(n)`
or `n_(p)=n_(i)sqrt(mu_(n)/mu_(p))`