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

Asked by vinitapahel123 | 1st Dec, 2015, 03:54: PM

Expert Answer:

Given:
Energy band gap of photodiode, Eg = 2.8 eV
Wavelength, λ = 600 nm = 6000 × 10-9 m
The energy of a signal is given by:
begin mathsize 14px style straight E equals hc over straight lambda end style
where, h = Planck's constant = 6.626 × 10-34 Js
c = Speed of light = 3 × 108 m/s
begin mathsize 14px style straight E equals hc over straight lambda equals fraction numerator 6.626 cross times 10 to the power of negative 34 end exponent cross times 3 cross times 10 to the power of 8 over denominator 6000 cross times 10 to the power of negative 9 end exponent end fraction equals 3.313 cross times 10 to the power of negative 20 end exponent space straight J therefore straight E equals 3.313 cross times 10 to the power of negative 20 end exponent space straight J because 1 space eV equals 1.6 cross times 10 to the power of negative 19 end exponent space straight J therefore straight E equals fraction numerator 3.313 cross times 10 to the power of negative 20 end exponent over denominator 1.6 cross times 10 to the power of negative 19 end exponent end fraction equals 0.207 space eV end style
The energy (E) of a signal of wavelength 6000 nm is 0.207 eV, which is less than 2.8 eV i.e, the energy band gap of a photodiode.
E < Eg.
Hence, the photodiode cannot detect the radiation of the given wavelength 6000 nm.

Answered by Faiza Lambe | 1st Dec, 2015, 04:25: PM