A photodiode is fabricated from a semiconductor with band gap of 2.8 eV. Can it detect a wavelength of 6000 nm? Justify.

Solution

Given,
Wavelength, λ = 6000 nm = 6 x 10^–6 m
Energy band gap, E_g = 2.8 eV

Using the formula for energy of a photon,

$E = \frac{hc}{λ}$

we have, $E = \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{6 \times 10^{- 6}} = 3.3 \times 10^{- 20} J$

$\Rightarrow$ $E = \frac{3.3 \times 10^{- 20}}{1.6 \times 10^{- 19}} = 0.206 eV$

As, the energy of the photon is less than energy band gap (E_g= 2.8 eV) of the semiconductor, so a wavelength of 6000 nm cannot be detected.

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Wave Optics

A photodiode is fabricated from a semiconductor with band gap of 2.8 eV. Can it detect a wavelength of 6000 nm? Justify.

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