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.
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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