Question

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

Answer 1

we have given,
wavelength, $\lambda$ = 6000nm = 6 × 10^-6 m
band gap, E = 2.8eV

we know the formula of energy of photon,
E = hc/$\lambda$

= 6.6 × 10^-34 × 3 × 10^8/(6 × 10^-6)
= 3.3 × 10^-20 J

hence, E =3.3 × 10^-20 J
now we can convert Joule in into eV.
so, E = 3.3 × 10^-20/(1.6 × 10^-19) = 0.206eV

as you can see that photon energy is less than band gap energy of semiconductor so , wavelength of 6000nm cannot be detected.

Answer 2

Hey !!

Energy corresponding to wavelength 6000 nm is

          E = hc / λ

= 6.6 × 10⁻³⁴ × 3 × 10⁸ / 6000 × 10⁻⁹ joule

             = 3.3 × 10⁻²⁰ J

= 3.3 × 10⁻²⁰ / 1.6 × 10⁻¹⁹

FINAL RESULT = 0.2 eV

The photon energy ( E = 0.2 eV )  of given wavelength is much less than the band gap ( Eg = 2.8 eV ) hence it cannot detect the given wavelength.

Good luck !!