Question

A bulb of 25 W emits yellow colour monochromatic light of wavelength 0.57 m. Calculate number of photon emmited per second

Answer 1

Power of bulb, P = 25 Watt = 25 Js^-1
Energy of one photon, E = hv = hc/λ.
Where ,
c = velocity of light in vacuum = 3 × 10⁸ m/s
h = Planck’s constant = 6.626 × 10^-34 Js
λ = 0.57 × 10^-6 m
Substituting the values, we get,
E = (6.626 × 10^-34) (3 × 10⁸)/ 0.57 × 10^-6
= 34.87 × 10^-20 J
Rate of emission of photon per second = 25/ 34.87 × 10^-20 = 7.169 × 10^-19 photon per second