Question

In a p-type Si sample, the hole concentration is \(2.25 \times 10^{15} \text{ cm}^{-3}\) and the intrinsic carrier concentration is \(1.5 \times 10^{10} \text{ cm}^{-3}\). The value of electron concentration will be

• A. Zero
• B. \(10^{10} \text{ cm}^{-3}\)
• C. \(10^5 \text{ cm}^{-3}\)
• D. \(1.5 \times 10^{25} \text{ cm}^{-3}\)

Hint:

The mass-action law, \(np = n_i^2\), is fundamental for finding minority carrier concentration in doped semiconductors under thermal equilibrium.

Answer 1

The Correct Option is C

Step 1: Recall the Mass-Action Law. For a semiconductor in thermal equilibrium, the product of the electron concentration (n) and the hole concentration (p) is a constant, equal to the square of the intrinsic carrier concentration (\(n_i\)). \[ n \cdot p = n_i^2 \]
Step 2: Identify the given values. Hole concentration (in p-type material, p \(\approx N_A\)): \( p = 2.25 \times 10^{15} \text{ cm}^{-3} \). Intrinsic carrier concentration: \( n_i = 1.5 \times 10^{10} \text{ cm}^{-3} \).
Step 3: Solve for the electron concentration (n). Rearrange the mass-action law formula: \[ n = \frac{n_i^2}{p} \] Substitute the given values: \[ n = \frac{(1.5 \times 10^{10})^2}{2.25 \times 10^{15}} = \frac{2.25 \times 10^{20}}{2.25 \times 10^{15}} \] \[ n = 1 \times 10^{(20-15)} = 10^5 \text{ cm}^{-3} \]