Question

A silicon crystal is doped with \(5\times 10^{17}\ \text{cm}^{-3}\) atoms of boron. The electron concentration \((n_{e})\) at room temperature is \((n_{i}=1.5\times 10^{10}\ \text{cm}^{-3})\)

• A. \(45.0\times 10^{2}\ \text{cm}^{-3}\)
• B. \(4.5\times 10^{10}\ \text{cm}^{-3}\)
• C. \(4.5\times 10^{2}\ \text{cm}^{-3}\)
• D. \(4.5\times 10^{7}\ \text{cm}^{-3}\)

Hint:

Boron makes the crystal p-type, so the hole density equals the doping level and the mass-action law gives the electron density.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
Boron is trivalent, so it makes p-type silicon. Hole concentration \(n_{h}\approx N_{A}=5\times 10^{17}\ \text{cm}^{-3}\).

Step 2: Mass-action law
\(n_{e}n_{h}=n_{i}^{2}\)

Step 3: Calculation
\(n_{e}=\dfrac{(1.5\times 10^{10})^{2}}{5\times 10^{17}}=\dfrac{2.25\times 10^{20}}{5\times 10^{17}}\)
\(n_{e}=4.5\times 10^{2}\ \text{cm}^{-3}\)

Final Answer:
The electron concentration is 4.5 times 10 to the power 2 per cubic centimetre. \[ \boxed{\text{(C) }4.5\times 10^{2}\ \text{cm}^{-3}} \]