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

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Payal Gupta | Contributor-Level 10

7 months ago

14.7 Energy band gap of the given photodiode, $E_{g}$ = 2.8 eV

Wavelength, $λ$ = 6000 nm = 6000 $* 10^{- 9}$ m

The energy of a signal is given by the relation,

$E$ = $\frac{h c}{λ}$

Where

h = Planck's constant = 6.626 $* 10^{- 34}$ Js

c = sped of light = 3 $* 10^{8}$ m/s

$E$ = $\frac{6.626 * 10^{- 34} * 3 * 10^{8}}{6000 * 10^{- 9}}$ = 3.313 $* 10^{- 20}$ J = $\frac{3.313 * 10^{- 20}}{1.6 * 10^{- 19}}$ eV = 0.207 eV

Therefore the energy of a signal of wavelength 6000 nm is 0.207 eV,

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