MHT CET · Physics · Semiconductors
A p-n junction photodiode is fabricated from semiconductor with a band gap of \(2.5 \mathrm{eV}\). It can detect a signal of wavelength
[Planck's constant \(=6.6 \times 10^{-34} \mathrm{Js}, \mathrm{c}=3 \times 10^8 \mathrm{~m} / \mathrm{s}, \mathrm{e}\) \(=1.6 \times 10^{-}\)\(\left.{}^{19} \mathrm{C}\right]\)
- A \(6000 \mathrm{~nm}\)
- B \(6000 Å\)
- C \(5000 Å\)
- D \(4000 \mathrm{~nm}\)
Answer & Solution
Correct Answer
(C) \(5000 Å\)
Step-by-step Solution
Detailed explanation
The wavelength corresponding to \(2.5 \mathrm{eV}\) is given by Energy \((\mathrm{eV})=\frac{12400}{\lambda(Å)}\)
\(
\therefore \lambda=\frac{12400}{2.5}=5000 Å
\)
To get detected, the signal should have energy greater than \(2.5 \mathrm{eV}\) or wavelength less than \(5000 Å\).
\(
\therefore \lambda=\frac{12400}{2.5}=5000 Å
\)
To get detected, the signal should have energy greater than \(2.5 \mathrm{eV}\) or wavelength less than \(5000 Å\).
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