KEAM 2019 Physics and Chemistry Paper

E=‌

hc

λ

E=‌

(6.626×10−34Js)(3.0×108)

550×10−9

E=0.036×10−17J
Each proton carries 0.036×10−17 J of energy.
P=IV=1A×100V
P=100W
A 100W source means that it outputs 100J of energy every one hour. So, in one second, 100×3600 J of energy is liberated.
Number of photons emitted = total energy output in one second /energy carried per photon
n=‌

100×3600

0.036×10−17

=1×1024