# A p-n photodiode is fabricated from a semiconductor with band gap of 2.8 eV.

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?

Solution:

Energy band gap of the given photodiode, Eg = 2.8 eV

Wavelength, λ = 6000 nm = 6000 × 10−9 m

The energy of a signal is given by the relation:

$E=\frac{h c}{\lambda}$

Where,

h = Planck’s constant

= 6.626 × 10−34 Js

c = Speed of light

= 3 × 108 m/s

$E=\frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{6000 \times 10^{-9}}$

= 3.313 × 10−20 J

But 1.6 × 10−19 J = 1 eV

E = 3.313 × 10−20 J

$=\frac{3.313 \times 10^{-20}}{1.6 \times 10^{-19}}=0.207 \mathrm{eV}$

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.