Power of the bulb = 100 watt = 100Js−1
Energy of one photon is E=hν=hc/λ
where, h=6.626×10−34Js,c=3×108ms−1
Given λ=400nm=400×10−9m
By putting the values, we get
E=6.626×10−34×3×108/(400×10−9)
E=4.969×10−19J
Number of photons emitted in 1 sec × energy of one photon = power
n×4.969×10−19=100
n=1004.969×10−19
n=2.012×1020photons per sec
So, option C is correct