A P-N photodiode is fabricated from a semiconductor with a band gap of 2.5 eV. It can detect a signal of wavelength :-

Options

A. 6000 Å
B. 6000 nmn
C. 4000 nm
D. 4000 Å

Answer

D

Explanation

Band gap = `E_g=2.5 eV= 2.5 xx 1.6 xx 10^(-19)` J
`=4xx10^(-19) J`
But E=hv =`"hc"/lambda=4xx10^(-19) J`
`therefore lambda=hc/E=(6.6xx10^(-34)xx3xx10^8)/(4xx10^(-19)) = 5xx10^(-7) m `
= 5000 Å
This is the threshold wavelength .
`therefore` It can detect a signal of wavelength `(lambda)` lt 5000 Å. Out of the four given wavelength, the only wavelength which is less than 5000 Å is 4000 Å.