For an n-type semiconductor, the relation between the number...

In an n-type semiconductor, pentavalent impurity atoms are added to the intrinsic semiconductor. Each impurity atom donates one free electron, significantly increasing the number density of electrons.

As a result, electrons become the majority charge carriers and holes become the minority charge carriers.

Therefore, the number density of electrons (n_e$) is much greater than the number density of holes (n_h$), which can be written as $n_h < < n_e$.

Answer: $n_h < < n_e$