Question

Assuming that the electron and hole masses equal the free electron mass, what is the intrinsic carrier density of diamond with a bandgap of 5.4 eV? Assume temperature 300 K.

Answer 1

The intrinsic carrier density of diamond is
`1.330*10^(-26)`.

Step-by-step explanation:

Given that,

Energy band gape = 5.4 eV

Temperature = 300 K

We need to calculate the intrinsic carrier density of diamond

Using formula of density

`n_(1)^2=N_(c)N_(v)e^{-(E_(g))/(kT)}`....(I)

We need to calculate
`N_(c)`

`N_(c)=2((2\pi m_(n)kT)/(h^2))^{(3)/(2)}`

Put the value into the formula

`N_(c)=2((2\pi*9.1*10^(-28)*300*1.3807*10^(-16))/((6.63*10^(-27))))^{(3)/(2)}`

`N_(c)=2.501*10^(19)\ cm^(-3)`

We need to calculate
`N_(v)`

`N_(v)=2((2\pi m_(n)kT)/(h^2))^{(3)/(2)}`

`N_(v)=2((2\pi*9.1*10^(-28)*300*1.3807*10^(-16))/((6.63*10^(-27))))^{(3)/(2)}`

`N_(v)=2.501*10^(19)\ cm^(-3)`

So.
`N_(c)=N_(v)`

Now, Put the value of
`N_(c)` and
`N_(v)` in equation (I)

`n_(i)^2=(2.501*10^(19))^2 e^{(-5.4)/(0.0259)}`

`n_(i)^2=1.7715080315*10^(-52)`

`n_(i)=1.330*10^(-26)`

Hence, The intrinsic carrier density of diamond is
`1.330*10^(-26)`.