A crystal of intrinsic silicon at room temperature has a carrier concentration of `1.6xx10^(16)m^(-3)` . If the donor concentration level is `4.8xx10^(20) m^(-3)` , then the concentration of holes in the semiconductor is

Options

A. `53xx10^(12)m^(-3)`
B. `4xx10^(11)m^(-3)`
C. `4xx10^(12)m^(-3)`
D. `5.3xx10^(11)m^(-3)`

Answer

D

Explanation

Given
`n_(1)=1.6xx10^(16)m^(-3)`
`n_(e)=4.8xx10^(20)m^(-3)`
`n_(h)=?`
The concentration of holes in the semiconductor is given by the expression
`n_(1)^(2)=n_(e)xxn_(h)`
`(1.6xx10^(16))^(2)=4.8xx10^(20)xxn_(h)`
`n_(h)=(2.56xx10^(32))/(4.8xx10^(20))=5.3xx10^(11)m^(-3)`