MHT CET · Physics · Semiconductors
In a pure silicon crystal electron-hole concentration is \(10^{16}\) per \(\mathrm{m}^3\) at 301 K . Now \(10^{21}\) atoms of phosphorus are added per cubic metre. The new hole concentration in silicon is (in per \(\mathrm{m}^3\) )
- A \(10^5\)
- B \(10^{11}\)
- C \(10^{19}\)
- D \(10^{21}\)
Answer & Solution
Correct Answer
(B) \(10^{11}\)
Step-by-step Solution
Detailed explanation
\( n \approx N_D = 10^{21} \, \mathrm{m}^{-3} \) \( p = \frac{n_i^2}{n} = \frac{(10^{16})^2}{10^{21}} = \frac{10^{32}}{10^{21}} = 10^{11} \, \mathrm{m}^{-3} \)
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