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)`.

Answer 1

`1"ppm"=1` part per million `=1/10^(6)`.
so no. of pentavalent As atoms doped in given Si
crystal, `=(5xx10^(28))/10^(6)=5xx10^(22)m^(-3)`. As one pentavalent
atoms donates one free electron to the crystal structure, so no. of free electrons in the given si
crystal, `n_(e)=5xx10^(22)m^(-3)`
Number of holes, `n_(h)=n_(i)^(2)/n_(e)=(1.5xx10^(16))^(2)/(5xx10^(22))`
`4.5xx10^(9)m^(-3)`