$ np =n_i^2 $\$ np =n_i^2 \$
Here, $ n= $\$ n= \$ Electron Charge Density ; $ p= $\$ p= \$ Hole Charge Density
And, $ n_i^2= $\$ n_i^2= \$ Intrinsic Charge Carrier Concentration
For n-type semiconductor, $ n>n_i $\$ n>n_i \$ so $ p\$ p<n_i \$
For p- type semiconductor, $ p>n_i $\$ p>n_i \$ so $ n
My book says that
Suppose we have an intrinsic semiconductor. It is known that, in a intrinsic semiconductor, $ n=p=n_i $\$ n=p=n_i \$
It is doped and turned into a n-type semicondctor.
My book says the number of holes in valence band/hole density, $p$\$p\$ decreases. My question is that does the hole density (minority charge carriers) in n-type semiconductor decreases only due to recombination process or is there any other reason other than the recombination process?
Now, if in the intrinsic semiconductor(sc), the hole density is $ p $\$ p \$ (which can be produced only by thermal agitation) and then that intrinsic sc is changed to n-type sc, the hole density, $ p $\$ p \$ will not change but the electron density $ n $\$ n \$ will increase by large amount($ n>>>p $\$ n>>>p \$) due to the presence of pentavalent impurity atoms and due to thermal agitation.
So the only way left with which the hole density, $ p $\$ p \$ can decrease is the process of recombination.
Suppose a crystal of intrinsic sc having 10 atoms(not possible but assume). 4 out of 10 atoms gets thermally excited and thus, 4 electron-hole pairs are created($n=4;p=4$\$n=4;p=4\$).
It is turned into a n-type semiconductor by doping it with 2 pentavalent impurity atoms. The 2 impurity atoms replaces the 2 host atoms so to maintain the total number of atoms, i.e., 10.
Thus, $ n=3+2=5 ; p=3 $\$ n=3+2=5 ; p=3 \$
So, hole density decreases from 4 to 3