Question

In a pure silicon crystal electron-hole concentration is $10^{16}$ per $\text{m}^3$ at $301\text{ K}$ . Now $10^{21}$ atoms of phosphorus are added per cubic metre. The new hole concentration in silicon is (in per $\text{m}^3$ )}

• A. $10^5$
• B. $10^{11}$
• C. $10^{19}$
• D. $10^{21}$

Hint:

For doped semiconductors, use: \[ np=n_i^2 \] After donor doping, \(n\) becomes very large and \(p\) becomes very small.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
In thermal equilibrium, the product of electron and hole concentrations is constant for a given temperature (Law of Mass Action).
Step 2: Key Formula or Approach:
Law of Mass Action: $n_e \times n_h = n_i^2$.
Step 3: Detailed Explanation:
Intrinsic concentration $n_i = 10^{16}\text{ m}^{-3}$.
Phosphorus is a pentavalent impurity (donor), so it increases the electron concentration $n_e \approx N_D = 10^{21}\text{ m}^{-3}$.
New hole concentration $n_h$ is:
\[ n_h = \frac{n_i^2}{n_e} = \frac{(10^{16})^2}{10^{21}} \]
\[ n_h = \frac{10^{32}}{10^{21}} = 10^{11}\text{ m}^{-3} \]
Step 4: Final Answer:
The new hole concentration is $10^{11}$ per $\text{m}^3$.