A photodiode is made froma semiconductor `In_(0.53) Gu_(0.47)A_(s), with E_(g)=0.73eV`.What is the maximum wavelength which it can detect ?`h=6.63xx10^(-34)J_(s)`

Explanation

The maximum wavelength`(lambda)`, which a photodiode , which a photodiode can detect corresponds to the energy `E_(g).So h_(c)/lambda=E_(g)`,or `lambda=h_(c)/E_(g)=(6.63xx10^(-34)xx (3xx10^(8)))/(0.73xx1.6xx10^(-19))`=`1703xx 10^(-9)m=1703nm`