COMEDK Exam 2014 Solved Paper

Power of the bulb =150W=150Js−1
As only 8% of the energy is emitted as light so, the total energy emitted per second
=

150J×8

100

=12J
Energy of one photon, E=hv=

hc

λ

=

(6.626×10−34Js)×(3×108ms−1)

6600×10−10m

=3.0118×10−19J
∴ Number of photons emitted per second
=

12J

3.0118×10−19J

=3.98×1019∞4.0×1019