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 \(=1242 \mathrm{eV} \mathrm{nm}\).)
- A \(407 \mathrm{~nm}\)
- B \(414 \mathrm{~nm}\)
- C \(207 \mathrm{~nm}\)
- D \(103.5 \mathrm{~nm}\)
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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