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Pure Si at 500K has equal number of electron (nₑ) and hole (nₕ) concentrations of 1.5 × 10¹⁶ m⁻³. Doping by indium increases nₕ to 4.5 × 10²² m⁻³. The doped semiconductor is of

1. n–type with electron concentration nₑ = 5 × 10²² m⁻³
2. p–type with electron concentration nₑ = 2.5 × 10¹⁰ m⁻³
3. n–type with electron concentration nₑ = 2.5 × 10¹⁰ m⁻³
4. p–type having electron concentration nₑ = 5 × 10⁻⁹ m⁻³

Correct answer: p–type with electron concentration nₑ = 2.5 × 10¹⁰ m⁻³

Solution

Indium is a trivalent impurity, so it introduces holes, making the semiconductor p-type. The product of electron and hole concentrations remains constant (nₑ × nₕ = constant). Using this, nₑ = (1.5 × 10¹⁶)² / 4.5 × 10²² = 2.5 × 10¹⁰ m⁻³.

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