We know that for each atom doped of Arsenic, one free electron is received. Similarly, for each atom doped of indium, a vacancy is created. So, the number of free electrons introduced by pentavalent impurity added
ne = NAs = 5 × 1022m–3
The number of holes introduced by trivalent impurity added
nh = NIn = 5 × 1020m–3
We know the relation, nenh = ni2 ...(i)
Now net electrons,
n′e = ne–nh = 5 × 1022 – 5 × 1020
= 4.95 × 1022m–3 ...(ii)
Now using equation (i), net holes
n′h =

ni2

n′e

=

(1.5×1016)2

4.95×1022

= 4.5 × 109m−3
So, n′e ≫ n′h , the material is of n-type.