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In an intrinsic semiconductor, $n_i = 10^{16} \, \text{m}^{-3}$. After doping with indium, the hole concentration becomes $n_h = 10^{20} \, \text{m}^{-3}$. The new electron concentration $n_e$ is

  1. A.$10^{12} \, \text{m}^{-3}$
  2. B.$10^{10} \, \text{m}^{-3}$
  3. C.$10^{8} \, \text{m}^{-3}$
  4. D.$10^{14} \, \text{m}^{-3}$
Show correct answer & step-by-step solution

Correct answer: A$10^{12} \, \text{m}^{-3}$

Solution

Using the mass action law , we find .

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