Question:
A pure silicon with $6 \times 10^{28}$ atoms is doped with arsenic of 1 ppm concentration. If the intrinsic carrier concentration is $1.2 \times 10^{16} m^{-3}$, then the number of holes in the doped silicon is}
A pure silicon with $6 \times 10^{28}$ atoms is doped with arsenic of 1 ppm concentration. If the intrinsic carrier concentration is $1.2 \times 10^{16} m^{-3}$, then the number of holes in the doped silicon is}
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For doped semiconductors:
\[
np=n_i^2
\]
This relation is called the law of mass action.
Updated On: Jun 17, 2026
- $2 \times 10^{12} m^{-3}$
- $1.2 \times 10^{16} m^{-3}$
- $5 \times 10^{12} m^{-3}$
- $2.4 \times 10^{9} m^{-3}$
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The Correct Option is D
Solution and Explanation
Concept:
For an n-type semiconductor,
\[
np=n_i^2
\]
where
\[
n=\text{electron concentration}
\]
\[
p=\text{hole concentration}
\]
\[
n_i=\text{intrinsic carrier concentration}
\]
Step 1: Find donor concentration.
\[ N_D = 6\times10^{28}\times10^{-6} \] \[ N_D=6\times10^{22}m^{-3} \] Therefore, \[ n\approx6\times10^{22}m^{-3} \]
Step 2: Use mass action law.
\[ np=n_i^2 \] \[ p= \frac{(1.2\times10^{16})^2} {6\times10^{22}} \] \[ p= 2.4\times10^9m^{-3} \] \[ \boxed{2.4\times10^9m^{-3}} \]
Step 1: Find donor concentration.
\[ N_D = 6\times10^{28}\times10^{-6} \] \[ N_D=6\times10^{22}m^{-3} \] Therefore, \[ n\approx6\times10^{22}m^{-3} \]
Step 2: Use mass action law.
\[ np=n_i^2 \] \[ p= \frac{(1.2\times10^{16})^2} {6\times10^{22}} \] \[ p= 2.4\times10^9m^{-3} \] \[ \boxed{2.4\times10^9m^{-3}} \]
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