Question

Three photo diodes $D_{1},$ $D_{2}$ and $D_{3}$ are made of semiconductors having bandgap of $2.5$eV, 2 eV and 3 eV, respectively. Which one will be able to detect light of wavelength 6000$\overset{o}{A}$?

• A. $D_{1}$
• B. $D_{2}$
• C. $D_{3}$
• D. $D_{1}$ and $D_{2}$ both

Answer 1

Answer: (B)

$D_{2}$

Answer 2

: Energy of incident photon,

$E = \frac{hc}{\lambda}$

$= \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{6 \times 10^{- 7} \times 1.6 \times 10^{- 19}} = 2.06eV$

The incident radiation can be detected by a photodiode if energy of incident photon is greater than the band gap.

As , $D_{2} = 2eV$

$\therefore D_{2}$will detect these radiations .