What is the concentration of holes in Si crystals having donor concentration of `1.6 xx 10^(24)//m^(3)`. When the intrinsic carrier concentration is `1.6 xx 10^(18)//m^(3)`? Calculate the ratio of electron holes concentration.

Explanation

Given, intrinsic carrier concentration, `n_(i)= 1.6 xx 10^(18)//m^(3)`
Donor concentration, `n_(D)= 1.6 xx 10^(24)//m^(3)`
Concentration of electron, `n_(e) ~~ n_(D)= 1.6 xx 10^(24)//m^(3)`
`therefore` Concentration of holes, `n_(h)= (n_(i)^(2))/(n_(e)) = ((1.6 xx 10^(18))^(2))/(1.6 xx 10^(24))`
`=1.6 xx 10^(12)//m^(3)`
Hence, ratio of electron to hole concentration is
`(n_(e))/(n_(h))= (1.6 xx 10^(24))/(1.6 xx 10^(12))= 10^(12)`