Assertion–Reason (Semiconductors) Assertion (A): For any semiconductor in thermal equilibrium, the carrier concentrations satisfy n p = n_i^2. Reason (R): A p-type semiconductor is obtained by adding a trivalent impurity to intrinsic material.

Difficulty: Easy

Correct Answer: Both A and R are true but R is not the correct explanation of A

Explanation:


Introduction / Context:
The mass–action law n p = ni^2 is a cornerstone of equilibrium semiconductor physics, relating electron and hole concentrations at a given temperature. Doping determines whether the material is p-type or n-type but does not invalidate the equilibrium relation itself. This question checks clarity on these two separate ideas.


Given Data / Assumptions:

  • Thermal equilibrium (no external excitation or strong injection).
  • n and p denote electron and hole concentrations; ni is intrinsic concentration.
  • p-type doping uses acceptor (trivalent) impurities in group-IV semiconductors like Si/Ge.


Concept / Approach:

Mass–action law: n p = ni^2 arises from detailed balance between generation and recombination processes and the position of Fermi level. Doping shifts individual values of n and p but their product remains fixed at equilibrium. Meanwhile, p-type doping via trivalent atoms (e.g., B in Si) creates acceptor states leading to hole majority carriers.


Step-by-Step Solution:

Evaluate A: n p = ni^2 → true in thermal equilibrium.Evaluate R: p-type via trivalent dopants → true statement.Causal link: R does not explain A; doping type does not derive mass–action law.Therefore, the correct choice is “both true, but R not the explanation”.


Verification / Alternative check:

Textbook derivations from Fermi statistics show n = NC exp [−(EC − EF)/kT], p = NV exp [−(EF − EV)/kT]; multiplying yields n p = NCNV exp [−(EC − EV)/kT] = ni^2.


Why Other Options Are Wrong:

  • A true, R false: R is factually correct.
  • R explains A: incorrect linkage.
  • A false: contradicts equilibrium semiconductor theory.


Common Pitfalls:

Believing that heavy doping breaks the mass–action law; at equilibrium it still holds (with bandgap narrowing caveats at extreme levels, but the principle remains).


Final Answer:

Both A and R are true but R is not the correct explanation of A

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