Question

Pure silicon at 300K has equal electron and hole concentration of \( 1.5 \times 10^{16} \) m\(^{-3}\). If the hole concentration increases to \( 3 \times 10^{22} \) m\(^{-3}\), then the electron concentration in the silicon is:

• A. \( 0.75 \times 10^9 \) m\(^{-3}\)
• B. \( 750 \) m\(^{-3}\)
• C. \( 75 \) m\(^{-3}\)
• D. \( 7.5 \times 10^9 \) m\(^{-3}\)

Hint:

In semiconductors, carrier concentration follows \( n_e = \frac{n_i^2}{n_h} \).

Answer 1

The Correct Option is D

Step 1: Apply Carrier Concentration Formula For intrinsic semiconductors: \[ n_i^2 = n_e n_h \] Step 2: Compute Electron Concentration \[ n_e = \frac{n_i^2}{n_h} \] \[ n_e = \frac{(1.5 \times 10^{16})^2}{3 \times 10^{22}} \] \[ n_e = \frac{2.25 \times 10^{32}}{3 \times 10^{22}} \] \[ n_e = 7.5 \times 10^9 \text{ m}^{-3} \] Thus, the correct answer is \( 7.5 \times 10^9 \) m\(^{-3}\). \bigskip

Answer 2

Given:
- Pure silicon at 300 K has intrinsic electron and hole concentrations:
\[ n_i = p_i = 1.5 \times 10^{16} \, \text{m}^{-3} \]

- Hole concentration increases to:
\[ p = 3 \times 10^{22} \, \text{m}^{-3} \]

We need to find the new electron concentration \( n \) in the silicon.

Step 1: Use the mass action law for semiconductors:
\[ np = n_i^2 \] where \( n \) is electron concentration, \( p \) is hole concentration, and \( n_i \) is intrinsic carrier concentration.

Step 2: Rearrange to solve for \( n \):
\[ n = \frac{n_i^2}{p} \]

Step 3: Substitute the values:
\[ n = \frac{(1.5 \times 10^{16})^2}{3 \times 10^{22}} = \frac{2.25 \times 10^{32}}{3 \times 10^{22}} = 0.75 \times 10^{10} = 7.5 \times 10^{9} \, \text{m}^{-3} \]

Therefore, the electron concentration in the silicon after the increase in hole concentration is:
\[ \boxed{7.5 \times 10^{9} \, \text{m}^{-3}} \]