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Question: Three photo diodes \(D_{1},\) \(D_{2}\) and \(D_{3}\) are made of semiconductors having bandgap of \...

Three photo diodes D1,D_{1}, D2D_{2} and D3D_{3} are made of semiconductors having bandgap of 2.52.5eV, 2 eV and 3 eV, respectively. Which one will be able to detect light of wavelength 6000Ao\overset{o}{A}?

A

D1D_{1}

B

D2D_{2}

C

D3D_{3}

D

D1D_{1} and D2D_{2} both

Answer

D2D_{2}

Explanation

Solution

: Energy of incident photon,

E=hcλE = \frac{hc}{\lambda}

=6.6×1034×3×1086×107×1.6×1019=2.06eV= \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 , D2=2eVD_{2} = 2eV

D2\therefore D_{2}will detect these radiations .