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

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" = ("hc")/lambda`

Where

h = Planck’s constant

= 6.626 × 10⁻³⁴ Js

c = Speed of light

= 3 × 10⁸ m/s

`"E" = (6.626 xx 10^(-34) xx 3 xx 10^8)/(6000 xx 10^(-9))`

= 3.313 × 10⁻²⁰ J

But 1.6 × 10⁻¹⁹ J = 1 eV

∴ E = 3.313 × 10⁻²⁰ J

`= (3.313 xx 10^(-20))/(1.6 xx 10^(-19)) = 0.207 " eV"`

The energy of a signal of wavelength 6000 nm is 0.207 eV, which is less than 2.8 eV − the energy band gap of a photodiode. Hence, the photodiode cannot detect the signal.