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A silicon bar is doped with donor impurities N_D = 2.25 × 10¹⁵ atoms/cm³. Given the intrinsic carrier concentration of silicon at T = 300 K is n_i = 1.5 × 10¹⁰ cm⁻³. Assuming complete impurity ionization, the equilibrium electron and hole concentrations are

1. n₀ = 1.5 × 10¹⁰ cm⁻³, p₀ = 1.5 × 10¹⁰ cm⁻³
2. n₀ = 1.5 × 10¹⁰ cm⁻³, p₀ = 1.5 × 10⁵ cm⁻³
3. n₀ = 2.25 × 10¹⁵ cm⁻³, p₀ = 1.5 × 10¹⁰ cm⁻³
4. n₀ = 2.25 × 10¹⁵ cm⁻³, p₀ = 1 × 10⁵ cm⁻³

Correct answer: n₀ = 2.25 × 10¹⁵ cm⁻³, p₀ = 1 × 10⁵ cm⁻³

Solution

The correct option reflects that the doping concentration of donor impurities significantly increases the electron concentration (n₀) to 2.25 × 10¹⁵ cm⁻³, while the hole concentration (p₀) decreases due to charge neutrality and is calculated using the mass action law, resulting in a much lower value of 1 × 10⁵ cm⁻³.

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