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

`E=hv`
` E=(hC)/lamda`
`E=(6.63xx10^(-34)xx3xx10^(8))/(600xx10^(-9)xx1.6xx10^(-19))`
`E=2.07 eV`
So diode `D_(3)` will not detect the light of wave length 600 nm.
(b) Consider the case of n type semi-conductor. The majority charge carrier density (n) is considerably larger than the minorty charge carrier hole density P (i.e. ` gt gt`P) on illumination, let the excess electrons and holes generated to `Deltan` and `DeltaP` respectively.
`n'=n+Deltan`
`P'=P+DeltaP`
`(Deltan)/nltltlt(DeltaP)/P" as n"ltltP`
Hence, this fractionl change in the majority charge carriers whould change in the majority charge carriers would be much less than that of the minority charge carriers.
Hence, photo diode are preferable used in teh reverse bias condition for measuring light intensity.