Saved Bookmarks
| 1. |
Energy gap in p - n photodiode is 2.8 eV. Can it detect a wavelength of 6000 nm ? Justify your answer. |
|
Answer» Solution :Here energy gap `E_(g)=2.8eV and lambda=6000 n =6000xx10^(-9)m=6xx10^(-6)m` `therefore"Energyof radiation PHOTON E"=(hc)/(elambda)eV=((6.63xx10^(-34))xx(3xx10^(8)))/((1.6xx10^(-19))xx(6xx10^(-6)))=0.207eV` As `E lt E_(g)`, the photodiode cannot detect the radiation. |
|