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JEE Mains · Physics · STD 12 - 14. Semicondutor electronics

The acceptor level of a p-type semiconductor is \(6 \mathrm{eV}\). The maximum wavelength of light which can create a hole would be _______. (Given hc \(=1242 \mathrm{eV} \mathrm{nm}\).)

  1. A \(407 \mathrm{~nm}\)
  2. B \(414 \mathrm{~nm}\)
  3. C \(207 \mathrm{~nm}\)
  4. D \(103.5 \mathrm{~nm}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(207 \mathrm{~nm}\)

Step-by-step Solution

Detailed explanation

\(\text { Energy }=\frac{hc}{\lambda}\) \(E=\frac{1240}{\lambda(\mathrm{nm})} \mathrm{eV}\) \(6=\frac{1240}{\lambda(\mathrm{nm})}\) \(\lambda=\frac{1240}{6}=207 \mathrm{~nm}\)
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