The number density of free electrons in the semiconductor is `10^(18)m^(-3)`. It is doped with a pentavalent impurity atoms of number density `10^(24)m^(-3)`, the number density of free electrons `m^(-3)` increases by a factor of

Options

A. `4//3`
B. 6
C. `10^(6)`
D. `10^(24)`

Answer

C

Explanation

The number density of free electrons in the doped semiconductor `=10^(18)+10^(24)~~10^(24) m^(-3)`.
Increase in number density by a factor
`=10^(24)//10^(18)=10^(6)`