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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 \(\mathrm{m}^3\) )

  1. A \(10^5\)
  2. B \(10^{11}\)
  3. C \(10^{19}\)
  4. D \(10^{21}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(10^{11}\)

Step-by-step Solution

Detailed explanation

\( n \approx N_D = 10^{21} \, \mathrm{m}^{-3} \) \( p = \frac{n_i^2}{n} = \frac{(10^{16})^2}{10^{21}} = \frac{10^{32}}{10^{21}} = 10^{11} \, \mathrm{m}^{-3} \)