Search any question & find its solution
Question:
Answered & Verified by Expert
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 $=6.6 \times 10^{-34} \mathrm{Js}, \mathrm{c}=3 \times 10^8 \mathrm{~m} / \mathrm{s}, \mathrm{e}=1.6 \times 10^{-}$ $\left.{ }^{19} \mathrm{C}\right]$
Options:
[Planck's constant $=6.6 \times 10^{-34} \mathrm{Js}, \mathrm{c}=3 \times 10^8 \mathrm{~m} / \mathrm{s}, \mathrm{e}=1.6 \times 10^{-}$ $\left.{ }^{19} \mathrm{C}\right]$
Solution:
1428 Upvotes
Verified Answer
The correct answer is:
$5000 Å$
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 $5000 Å$.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.