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

number density of free electrons is `10^(18)+10^(24) ~=10^(24)` increase by factor `(10^(24))/10^(18)=10^6`.