Question:
A pure semiconductor crystal has \(8\times10^{28}\ \text{atoms m}^{-3}\). It is doped by \(2\ \text{ppm}\) concentration of pentavalent atoms. The number of holes formed in the semiconductor crystal is
\[
(\text{Intrinsic carrier concentration, } n_i=1\times10^{16}\ \text{m}^{-3})
\]
A pure semiconductor crystal has \(8\times10^{28}\ \text{atoms m}^{-3}\). It is doped by \(2\ \text{ppm}\) concentration of pentavalent atoms. The number of holes formed in the semiconductor crystal is
\[ (\text{Intrinsic carrier concentration, } n_i=1\times10^{16}\ \text{m}^{-3}) \]
\[ (\text{Intrinsic carrier concentration, } n_i=1\times10^{16}\ \text{m}^{-3}) \]
Show Hint
For doped semiconductors, use the mass action law:
\[
np=n_i^2.
\]
For an \(n\)-type semiconductor,
\[
n\approx N_D,
\]
so the hole concentration is
\[
p=\frac{n_i^2}{N_D}.
\]
Updated On: Jun 26, 2026
- \(4.5\times10^9\ \text{m}^{-3}\)
- \(6.25\times10^8\ \text{m}^{-3}\)
- \(2.5\times10^9\ \text{m}^{-3}\)
- \(1.25\times10^8\ \text{m}^{-3}\)
Show Solution
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The Correct Option is B
Solution and Explanation
Step 1: Identify the type of semiconductor.
The semiconductor is doped with pentavalent impurity atoms.
Pentavalent impurities are donor impurities, so the semiconductor becomes an \(n\)-type semiconductor.
In an \(n\)-type semiconductor, electrons are majority carriers and holes are minority carriers.
Step 2: Calculate donor concentration.
Total number of atoms in the pure semiconductor is \[ 8\times10^{28}\ \text{m}^{-3}. \] Doping concentration is \[ 2\ \text{ppm}=2\times10^{-6}. \] Therefore, donor concentration is \[ N_D=(8\times10^{28})(2\times10^{-6}). \] \[ N_D=16\times10^{22}. \] \[ N_D=1.6\times10^{23}\ \text{m}^{-3}. \] For an \(n\)-type semiconductor, \[ n\approx N_D. \] So, \[ n=1.6\times10^{23}\ \text{m}^{-3}. \]
Step 3: Use mass action law.
For a semiconductor, \[ np=n_i^2, \] where \(n\) is electron concentration, \(p\) is hole concentration, and \(n_i\) is intrinsic carrier concentration.
Therefore, \[ p=\frac{n_i^2}{n}. \] Given, \[ n_i=1\times10^{16}\ \text{m}^{-3}. \] Thus, \[ p=\frac{(1\times10^{16})^2}{1.6\times10^{23}}. \] \[ p=\frac{1\times10^{32}}{1.6\times10^{23}}. \] \[ p=0.625\times10^9. \] \[ p=6.25\times10^8\ \text{m}^{-3}. \]
Step 4: Final conclusion.
Therefore, the number of holes formed in the semiconductor crystal is \[ \boxed{6.25\times10^8\ \text{m}^{-3}} \] Hence, the correct option is \[ \boxed{(2)} \]
The semiconductor is doped with pentavalent impurity atoms.
Pentavalent impurities are donor impurities, so the semiconductor becomes an \(n\)-type semiconductor.
In an \(n\)-type semiconductor, electrons are majority carriers and holes are minority carriers.
Step 2: Calculate donor concentration.
Total number of atoms in the pure semiconductor is \[ 8\times10^{28}\ \text{m}^{-3}. \] Doping concentration is \[ 2\ \text{ppm}=2\times10^{-6}. \] Therefore, donor concentration is \[ N_D=(8\times10^{28})(2\times10^{-6}). \] \[ N_D=16\times10^{22}. \] \[ N_D=1.6\times10^{23}\ \text{m}^{-3}. \] For an \(n\)-type semiconductor, \[ n\approx N_D. \] So, \[ n=1.6\times10^{23}\ \text{m}^{-3}. \]
Step 3: Use mass action law.
For a semiconductor, \[ np=n_i^2, \] where \(n\) is electron concentration, \(p\) is hole concentration, and \(n_i\) is intrinsic carrier concentration.
Therefore, \[ p=\frac{n_i^2}{n}. \] Given, \[ n_i=1\times10^{16}\ \text{m}^{-3}. \] Thus, \[ p=\frac{(1\times10^{16})^2}{1.6\times10^{23}}. \] \[ p=\frac{1\times10^{32}}{1.6\times10^{23}}. \] \[ p=0.625\times10^9. \] \[ p=6.25\times10^8\ \text{m}^{-3}. \]
Step 4: Final conclusion.
Therefore, the number of holes formed in the semiconductor crystal is \[ \boxed{6.25\times10^8\ \text{m}^{-3}} \] Hence, the correct option is \[ \boxed{(2)} \]
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