Question:
A pure silicon crystal at temperature \(300\) K has electron and hole concentration (\(n_i\)) \(10^{16}\) per m\(^3\) each and \(10^{21}\) phosphorus atoms (\(n_e\)) are added per cubic metre. The new hole concentration in silicon crystal is
A pure silicon crystal at temperature \(300\) K has electron and hole concentration (\(n_i\)) \(10^{16}\) per m\(^3\) each and \(10^{21}\) phosphorus atoms (\(n_e\)) are added per cubic metre. The new hole concentration in silicon crystal is
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Use the mass action law \(n_en_h=n_i^2\).
Updated On: Oct 1, 2026
- \(10^{19}\) per m\(^3\)
- \(10^{21}\) per m\(^3\)
- \(10^5\) per m\(^3\)
- \(10^{11}\) per m\(^3\)
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept:
In a doped semiconductor at a fixed temperature, \(n_en_h = n_i^2\).
Step 2: Calculate:
Phosphorus is a donor, so the electron concentration is about the donor density: \(n_e\approx10^{21}\) m\(^{-3}\). Thermally generated electrons \(10^{16}\) are negligible in comparison.
\[ n_h = \frac{n_i^2}{n_e} = \frac{(10^{16})^2}{10^{21}} = 10^{11}\ \text{m}^{-3} \]
Final Answer:
The new hole concentration is \(10^{11}\) per m\(^3\), option (D). \[ \boxed{10^{11}\ \text{m}^{-3}} \]
In a doped semiconductor at a fixed temperature, \(n_en_h = n_i^2\).
Step 2: Calculate:
Phosphorus is a donor, so the electron concentration is about the donor density: \(n_e\approx10^{21}\) m\(^{-3}\). Thermally generated electrons \(10^{16}\) are negligible in comparison.
\[ n_h = \frac{n_i^2}{n_e} = \frac{(10^{16})^2}{10^{21}} = 10^{11}\ \text{m}^{-3} \]
Final Answer:
The new hole concentration is \(10^{11}\) per m\(^3\), option (D). \[ \boxed{10^{11}\ \text{m}^{-3}} \]
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