Question

In a pure silicon, number of electrons and holes per unit volume are $1.6 \times 10^{16}\ \text{m}^{-3}$. If silicon is doped with Boron in a way that on doping hole density increases to $4 \times 10^{22}\ \text{m}^{-3}$. Then electron density in doped semiconductor will be

• A. $6.4 \times 10^{-9}\ \text{m}^{-3}$
• B. $6.4 \times 10^{9}\ \text{m}^{-3}$
• C. $6.4 \times 10^{-10}\ \text{m}^{-3}$
• D. $6.4 \times 10^{10}\ \text{m}^{-3}$

Hint:

The Law of Mass Action states that the product of carrier concentrations in a semiconductor remains constant at a given temperature, even after doping.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We are dealing with a doped semiconductor. We need to find the new electron density ($n_e$) given the intrinsic carrier concentration ($n_i$) and the new hole density ($n_h$).

Step 2: Key Formula or Approach:
Use the Law of Mass Action for semiconductors: $n_e \cdot n_h = n_i^2$.

Step 3: Detailed Explanation:
Given:
$n_i = 1.6 \times 10^{16}\ \text{m}^{-3}$
$n_h = 4 \times 10^{22}\ \text{m}^{-3}$
Using $n_e = \frac{n_i^2}{n_h}$:
$n_e = \frac{(1.6 \times 10^{16})^2}{4 \times 10^{22}}$
$n_e = \frac{2.56 \times 10^{32}}{4 \times 10^{22}}$
$n_e = \frac{2.56}{4} \times 10^{32-22} = 0.64 \times 10^{10} = 6.4 \times 10^{9}\ \text{m}^{-3}$.

Step 4: Final Answer:
The electron density is $6.4 \times 10^{9}\ \text{m}^{-3}$, which corresponds to option (B).