In a semiconductor, the number density of intrinsic charge carries at 27°C is 1.5 × 10¹⁶/m³. If the semiconductor is doped with impurity atom, the hole density increases to 4.5 × 10²²/m³. The electron density in the doped semiconductor is ___________ × 10⁹/m³.

Solution

Physics sir · 3 steps

Doping changes n_e and n_h, but their product stays fixed at n_i squared. Use that and you are done.

1. Mass action law. At a fixed temperature, the product of electron and hole densities in a semiconductor equals the square of the intrinsic carrier density, doped or not.

   nₑ nₕ = nᵢ²

2. Put in numbers. nᵢ² = (1.5 × 10¹⁶)² = 2.25 × 10³² m⁻⁶. Then nₑ = (2.25 × 10³²)/(4.5 × 10²²) = 0.5 × 10¹⁰ = 5 × 10⁹ m⁻³.
3. So. The electron density is 5 × 10⁹ m⁻³, so the blank is 5.

The common slip: 5000000000. Writing the full density gives 5 000 000 000, but the blank already carries × 10⁹, so the entry is just 5.

The idea to keep. Doping raises one carrier type and lowers the other, but at a fixed temperature nₑ nₕ = nᵢ² always holds. Square nᵢ and divide by the carrier you know.

Answer from the NTA final answer key. Solution by PrepWiser.