A crystal of intrinsic silicon at room temperature has a carrier concentration of `1.6 xx 10^(6)//m^(3)`. If the donor concentration level is `4.8 xx 10^(20)//m^(3)`, then the concentration of holes in the semiconductor is

Options

A. `53 xx 10^(12)//m^(3)`
B. `4 xx 10^(11)//m^(3)`
C. `4 xx 10^(12)//m^(3)`
D. `5.3 xx 10^(11)//m^(3)`

Answer

D

Explanation

The connection of holes in the semiconductor is,
`n^(2)_(i) = n_(e) xx n_(h)`
`(1.6 xx 10^(16))^(2) = 4.8 xx 10^(20) xx n_(h)`
`n_(h) = 2.56 xx 10^(32)/4.8 xx 10^(20) = 0.533 xx 10^(12) = 5.33 xx 10^(11)//m^(3)`