Pure Si at 300K has hole and electron densities are 1.5 x 10^16 m^-3. ...
Solution:Given: Hole density (n(h)) in pure Si = 1.5 x 10^16 m^-3
Doped Si hole density (n(h)) = 4.5 x 10^22 m^-3
Electron density (n(e)) in doped Si = ?
Calculations:It is given that the Si is doped with impurity, which means that either p-type or n-type semiconductor is formed.
As the hole density (n(h)) is increased, it indicates that the impurity added is a p-type impurity.
The relation between electron density (n(e)) and hole density (n(h)) for a p-type semiconductor is given as:
n(e) * n(h) = ni^2
Where, ni is the intrinsic carrier concentration, which is equal to 1.5 x 10^10 m^-3 for Si at 300K.
Substituting the given values, we get:
n(e) = ni^2 / n(h) = (1.5 x 10^10 m^-3)^2 / (4.5 x 10^22 m^-3) = 5 x 10^9 m^-3
Answer:Therefore, the electron density in the doped Si is 5 x 10^9 m^-3.