Three photodiodes `D_1 , D_2 ` and `D_3` are made of semiconductors having
band gaps of ` 2.5 eV , 2 eV` and 3 eV , respectively . Which one will be able to detect light of wavelength ` 6000 Å ` ?

Explanation

Energy of incident light is given by
`E= hv - (hc)/( lambda) = (6.63 xx 10^(-34) xx 3 xx 10^(8))/( 6,000 xx 10^(-10) xx (1.6 xx 10^(-6))) = 2.07 eV`
So , only diode `D_(2)` will detect this radiation because the energy of incident photon should be greater than the band gap of the diode.