Question

An intrinsic semiconductor has equal concentrations of hole and electron which is equal to \(4 \times 10^{8}\,\text{m}^{-3}\). During its conversion to extrinsic semiconductor, concentration of hole increases to \(8 \times 10^{10}\,\text{m}^{-3}\). The new electron concentration is:

• A. \(4 \times 10^{8}\,\text{m}^{-3}\)
• B. \(2 \times 10^{8}\,\text{m}^{-3}\)
• C. \(2 \times 10^{6}\,\text{m}^{-3}\)
• D. \(4 \times 10^{10}\,\text{m}^{-3}\)

Hint:

Always use \(np = n_i^2\) (mass action law) in semiconductor problems. If one carrier increases, the other decreases.

Answer 1

The Correct Option is C

Step 1: Use intrinsic carrier concentration relation.
For intrinsic semiconductor:
\[ n_i = p_i = 4 \times 10^{8} \]

Step 2: Use mass action law.
\[ np = n_i^2 \]

Step 3: Substitute intrinsic value.
\[ n_i^2 = (4 \times 10^{8})^2 \]
\[ n_i^2 = 16 \times 10^{16} \]

Step 4: Use new hole concentration.
\[ p = 8 \times 10^{10} \]

Step 5: Find new electron concentration.
\[ n = \frac{n_i^2}{p} \]
\[ n = \frac{16 \times 10^{16}}{8 \times 10^{10}} \]

Step 6: Simplify.
\[ n = 2 \times 10^{6}\,\text{m}^{-3} \]

Step 7: Final answer.
\[ \boxed{2 \times 10^{6}\,\text{m}^{-3}} \]