Minority-carrier density in an n-type semiconductor (use mass-action law) An n-type semiconductor at 300 K has electron concentration n = 6.25 × 10^18 cm^-3. The intrinsic carrier concentration is n_i = 2.5 × 10^13 cm^-3. What is the hole concentration p?

Introduction / Context:Carrier concentrations in semiconductors at equilibrium obey the mass-action law. Given majority carrier density in an n-type sample and intrinsic carrier concentration, you can compute the minority carrier density (holes) directly. This is fundamental for diode and transistor analysis.

Given Data / Assumptions:

• Electron concentration n = 6.25 × 10^18 cm^-3.
• Intrinsic concentration n_i = 2.5 × 10^13 cm^-3 at 300 K.
• Thermal equilibrium (no significant generation-recombination transients).

Concept / Approach:

The mass-action law states n * p = n_i^2 for a homogeneous semiconductor at equilibrium. Solve for p = n_i^2 / n. Keep careful track of powers of ten to avoid arithmetic mistakes.

Step-by-Step Solution:

Verification / Alternative check:

Order-of-magnitude check: heavy n-type doping (10^18) should force holes down to a very low value (10^8), consistent with n * p = 10^26.

Why Other Options Are Wrong:

• 10^6 or 10^10 or 10^12 cm^-3 do not satisfy n * p = n_i^2 with the given n and n_i.

Common Pitfalls:

Squaring 2.5 incorrectly; misplacing exponents; forgetting that units of n and p are per cm^3.

Final Answer:

10^8 cm^-3