Consider a n-type semiconductor in which nₑ and nₕ are number of electrons and holes, respectively.
(A) Holes are minority carriers
(B) The dopant is a pentavalent atom
(C) nₑ nₕ ≠ nᵢ²
(where nᵢ is number of electrons or holes in semiconductor when it is intrinsic form)
(D) nₑ nₕ ≥ nᵢ²
(E) The holes are not generated due to the donors
Choose the correct answer from the options given below :

Solution

Physics sir · 3 steps

Doping changes how many electrons and holes there are, but their product stays fixed at nᵢ².

1. What n-type means. A pentavalent atom such as P or As has 5 valence electrons. Four go into bonds and the fifth is donated to the crystal, so electrons become the majority carriers and holes the minority carriers.
2. Mass action law. In thermal equilibrium, doping does not change the product of the carrier concentrations. Extra electrons recombine with some holes, so nₕ drops by just enough to keep the product constant.

   nₑ nₕ = nᵢ²

   Statement	Correct?	Reason
   (A) Holes are minority carriers	Yes	Donors add electrons, so nₑ ≫ nₕ
   (B) Dopant is pentavalent	Yes	Its fifth electron is donated
   (C) nₑ nₕ ≠ nᵢ²	No	The mass action law makes them equal
   (D) nₑ nₕ ≥ nᵢ²	No	The product cannot exceed nᵢ². It is fixed at exactly nᵢ²
   (E) Donors do not generate holes	Yes	A donor gives an electron without leaving a hole in a bond. Holes come only from thermal bond breaking

3. So. The correct statements are (A), (B) and (E), which is option (a).

The idea to keep. In any doped semiconductor at equilibrium, nₑ nₕ = nᵢ². Donors add free electrons only and leave no holes behind, so holes come only from thermal generation.

Answer from the NTA final answer key. Solution by PrepWiser.