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

- (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).