Q. A $p$-$n$ photodiode is fabricated from a semiconductor with a band gap of $2.5\ eV$. It can detect a signal of wavelength
A) 6000 $\mathring{\mathrm{A}}$
B) $4000\ nm$
C) $6000\ nm$
D) $4000\ \mathring{\mathrm{A}}$
Explanation
(d) $E = \frac{hc}{\lambda} \Rightarrow \lambda = \frac{12400}{E(ev)} = \frac{12400}{2.5} = 4960\mathring{\mathrm{A}}$
This is the maximum wavelength i.e., the signals having wavelength greater than this value are not detected by photodiode. Hence correct option is (d)
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.