Question

Suppose a pure Si-crystal has `5xx10^(28) "atoms" m^(-3)`. It is doped by 1 ppm concentration of pentavalent As. Calculate the number of electrons and holes. Give that `n_(i)=1.5xx10^(16)m^(-3)`.
A. `4.5xx10^9 m^(-3)`
B. `4.5xx10^6 m^(-3)`
C. `2.5xx10^9 m^(-3)`
D. `2.5xx10^6 m^(-3)`

Answer 1

Correct Answer - A
The electron and holes concentration in a semiconductor in thermal equilibrium is
`n_e n_h=n_i^2`
or `n_h=n_i^2/n_e`
Here , `n_i=1.5xx10^16 m^(-3) , n_e=5xx10^22 m^(-3)`
`therefore n_h=((1.5xx10^16)^2)/((5xx10^22))=(2.25xx10^32)/(5xx10^22)=4.5xx10^9 m^(-3)`