Question

At 300 K, both electron and hole density in an intrinsic silicon crystal is $ 15 \times 10^{15} \, \text{m}^{-3} $. When it is doped with indium, the hole density becomes $ 4.5 \times 10^{22} \, \text{m}^{-3} $. The extrinsic electron density will be

• A. \( 6 \times 10^9 \, \text{m}^{-1} \)
• B. \( 5 \times 10^9 \, \text{m}^{-3} \)
• C. \( 4 \times 10^9 \, \text{m}^{-3} \)
• D. \( 3 \times 10^9 \, \text{m}^{-3} \)

Hint:

In semiconductor physics, use the product of electron and hole densities to find unknown densities in extrinsic semiconductors.

Answer 1

The Correct Option is B

Step 1: Use the relationship between electron and hole densities.
In an intrinsic semiconductor, the product of the electron and hole densities is constant and equal to the square of the intrinsic carrier concentration: \[ n_i^2 = p_i \times n_i \] where: \( n_i \) is the intrinsic carrier concentration, \( p_i \) is the hole density, \( n_i \) is the electron density.
Step 2: Find the extrinsic electron density.
For an extrinsic semiconductor, the product of the electron density and hole density is still related by: \[ n_{\text{extrinsic}} \times p_{\text{extrinsic}} = n_i^2 \] We are given:
\( p_{\text{extrinsic}} = 4.5 \times 10^{22} \, \text{m}^{-3} \), \( p_{\text{intrinsic}} = 15 \times 10^{15} \, \text{m}^{-3} \). Thus, the extrinsic electron density is: \[ n_{\text{extrinsic}} = \frac{n_i^2}{p_{\text{extrinsic}}} = \frac{(15 \times 10^{15})^2}{4.5 \times 10^{22}} = 5 \times 10^9 \, \text{m}^{-3} \]
Step 3: Conclusion.
Thus, the extrinsic electron density is \( 5 \times 10^9 \, \text{m}^{-3} \).
Conclusion:
The correct answer is (B) \( 5 \times 10^9 \, \text{m}^{-3} \).