Question

The electron density of intrinsic semiconductor at room temperature is \( 10^{16} \,\text{m}^{-3} \). When doped with a trivalent impurity, the electron density is decreased to \( 10^{14} \,\text{m}^{-3} \) at the same temperature. The majority carrier density is

• A. $10^{16} \,\text{m}^{-3}$
• B. $10^{18} \,\text{m}^{-3}$
• C. $10^{21} \,\text{m}^{-3}$
• D. $10^{20} \,\text{m}^{-3}$
• E. $10^{19} \,\text{m}^{-3}$

Hint:

Always use $np = n_i^2$ to find unknown carrier concentration.

Answer 1

The Correct Option is B

Concept: Mass action law in semiconductors
\[ n p = n_i^2 \] where:
• $n$ = electron concentration
• $p$ = hole concentration
• $n_i$ = intrinsic carrier concentration ---

Step 1: Given intrinsic carrier density \[ n_i = 10^{16} \,\text{m}^{-3} \] ---

Step 2: After doping (p-type) \[ n = 10^{14} \,\text{m}^{-3} \] ---

Step 3: Apply mass action law \[ p = \frac{n_i^2}{n} \] \[ p = \frac{(10^{16})^2}{10^{14}} = \frac{10^{32}}{10^{14}} = 10^{18} \] ---

Step 4: Identify majority carriers
• Trivalent impurity → p-type semiconductor
• Holes are majority carriers --- Final Answer: \[ \boxed{10^{18} \,\text{m}^{-3}} \]