ExamBro
ExamBro
JEE Mains · Physics · STD 12 - Semicondutor electronics
  • JEE Mains 2024
  • Hard
  • MCQ

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 \(=1242 \mathrm{eV} \mathrm{nm}\).)

Pick an option to check your answer.
Verified Solution

Answer & Solution

Correct Answer

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

Step-by-step Solution

Detailed explanation

\(\text { Energy }=\frac{hc}{\lambda}\)

\(E=\frac{1240}{\lambda(\mathrm{nm})} \mathrm{eV}\)

\(6=\frac{1240}{\lambda(\mathrm{nm})}\)

\(\lambda=\frac{1240}{6}=207 \mathrm{~nm}\)

Same subject
Explore more questions on app