Question

Consider a n-type semiconductor in which $ n_e $ and $ n_h $ are the number of electrons and holes, respectively.

• A. Holes are minority carriers
• B. The dopant is a pentavalent atom
• C. \( n_e n_h = n_i^2 \) for intrinsic semiconductor
• D. \( n_e \gg n_h \) for extrinsic semiconductor The correct answer from the options given below is:

Hint:

In n-type semiconductors, electrons are the majority carriers, and holes are the minority carriers. The product of electron and hole concentrations is related to the intrinsic carrier concentration in an intrinsic semiconductor.

Answer 1

The Correct Option is C

To understand the correct answer, let's evaluate each statement given in the options with respect to an n-type semiconductor:

1. Holes are minority carriers: In an n-type semiconductor, electrons are the majority charge carriers because they are supplied by the pentavalent dopant atoms, typically Phosphorus or Arsenic. Consequently, the number of electrons \(n_e\) is much larger than the number of holes \(n_h\). Therefore, holes indeed are minority carriers. However, this option is incorrect as per the given correct answer.
2. The dopant is a pentavalent atom: In n-type semiconductors, the dopants are typically pentavalent, meaning they have five valence electrons. Examples include Phosphorus, Arsenic, or Antimony. This statement is true for n-type semiconductors, but this is not the correct answer as per the problem.
3. \( n_e n_h = n_i^2 \) for intrinsic semiconductor: This formula represents the mass action law in semiconductors. Here, \(n_i\) is the intrinsic carrier concentration. This equation is valid for any semiconductor condition, regardless of doping. In an intrinsic semiconductor, \(n_e = n_h = n_i\). Hence, this statement is always true and matches the given correct answer.
4. \( n_e \gg n_h \) for extrinsic semiconductor: In an n-type semiconductor, which is an example of an extrinsic semiconductor, the electron concentration \(n_e\) is indeed much greater than the hole concentration \(n_h\), due to additional electrons from dopant atoms. Although this statement holds true in practice, this is not the correct option as per the problem statement.

Based on this analysis, the correct answer is indeed: \( n_e n_h = n_i^2 \) for intrinsic semiconductor. This statement is based on the fundamental principle of semiconductor physics and holds true under equilibrium conditions for any type of semiconductor, whether intrinsic (pure) or extrinsic (doped).

Answer 2

- (A) Holes are minority carriers: In an n-type semiconductor, the electrons are the majority carriers, and the holes are the minority carriers. 
- (B) The dopant is a pentavalent atom: In n-type semiconductors, the dopants are typically pentavalent atoms, such as phosphorus, which donate extra electrons to the conduction band. 
- (C) \( n_e n_h = n_i^2 \) for intrinsic semiconductor: For an intrinsic semiconductor, the product of the electron and hole concentrations is equal to the square of the intrinsic carrier concentration, i.e., \( n_e n_h = n_i^2 \). 
- (D) \( n_e \gg n_h \) for extrinsic semiconductor: This is true for n-type semiconductors, where the electron concentration is much greater than the hole concentration. 
Thus, the correct answer is (3).