Question

Find the number of photons emitted per second by a 25-watt bulb of monochromatic light of wavelength 6000A. What is the photoelectric current assuming 3% efficiency for photoelectric effect?

Answer 1

Explanation:

Pin=25W,λ=6600×10−10mnhv=P

⇒Numberofphotonsemitted∖sec,

n=Phcλ=Pλhc=25×6600×10−106.64×10−34×3×108

=8.28×1019=253×10193%ofemittedphotonsareproducingcurrent

∴I=3100×ne=3100×253×1019×1.6×10−19=0.4A

Questions from BITSAT 2016