Question

A heavily doped n-type semiconductor has the following data:
Hole-electron mobility ratio : $0.4$
Doping concentration: $4.2\times {10}^{8}atoms/m^3$
Intrinsic concentration: $1.5\times {10}^{4}atoms/m^3$
The ratio of conductance of the n-type semiconductor to that of the intrinsic semiconductor of same material and at the same temperature is given by

Answer 1

Ans : (d)

For n-type semiconductor ${\sigma }_n=nq{\mu }_n$
For the ratio of conductance of the n-type semiconductor to that of the intrinsic semiconductor of same material
$\frac{{\sigma }_n}{{\sigma }_i}=\frac{n{\mu }_n}{n_i({\mu }_n+{\mu }_p)}$
$=\frac{4.2\times {10}^8\times {\mu }_n}{1.5\times {10}^4\times {\mu }_n(1+\frac{{\mu }_p}{{\mu }_n})}$
$=\frac{4.2\times {10}^8}{1.5\times {10}^4\times 1.4}=2\times {10}^4$