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:

Use the mass action law: the product of electron and hole concentrations stays equal to the square of the intrinsic concentration.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Phosphorus is a donor, so it makes the crystal n-type. The law \(n_e n_h = n_i^2\) holds.

Step 2: Electron Concentration
Each phosphorus atom gives one free electron, so \(n_e \approx 10^{21}\ \text{m}^{-3}\) (much larger than \(10^{16}\)).

Step 3: Calculation
\(n_h = \dfrac{n_i^2}{n_e} = \dfrac{(10^{16})^2}{10^{21}} = 10^{11}\ \text{m}^{-3}\).

Step 4: Compare the options
The hole concentration is \(10^{11}\), the second option.

Final Answer:
The new hole concentration is 10 to the power 11 per cubic metre. \[ \boxed{\text{(B) }10^{11}} \]