To determine the correct answer, each statement regarding an n-type semiconductor is evaluated:
- Holes are minority carriers: In an n-type semiconductor, electrons are the majority carriers due to pentavalent dopants (e.g., Phosphorus, Arsenic). Therefore, the electron concentration \(n_e\) significantly exceeds the hole concentration \(n_h\). This statement correctly identifies holes as minority carriers. However, it is not the given correct answer.
- The dopant is a pentavalent atom: N-type semiconductors are doped with pentavalent atoms, possessing five valence electrons, such as Phosphorus, Arsenic, or Antimony. This statement is factual for n-type semiconductors but is not the correct answer according to the problem.
- \( n_e n_h = n_i^2 \) for intrinsic semiconductor: This equation represents the mass action law for semiconductors, where \(n_i\) is the intrinsic carrier concentration. It is universally applicable to all semiconductor conditions, irrespective of doping. In an intrinsic semiconductor, \(n_e = n_h = n_i\). This statement is always true and aligns with the provided correct answer.
- \( n_e \gg n_h \) for extrinsic semiconductor: An n-type semiconductor is an extrinsic type. Here, the electron concentration \(n_e\) is substantially greater than the hole concentration \(n_h\) due to the added electrons from dopants. While practically true, this is not the designated correct option in the problem.
Consequently, the correct answer is: \( n_e n_h = n_i^2 \) for intrinsic semiconductor. This principle is a foundational concept in semiconductor physics and remains valid under equilibrium conditions for all semiconductors, whether intrinsic or extrinsic.