A 25-watt bulb emits monochromatic yellow light of wavelength of 0.57μm. Calculate the rate of emission of quanta per second.
Given:
Power of the bulb, P = 25 W
Wavelength of light emitted, λ = 0.57μm
By Planck’s relation we have,
Energy, E = h×v
But we know v = [c] / [λ]
Where
c = Speed of Light
v= Frequency
λ = Wavelength
So, E = hc /λ
= [[6.626×10-34] × [3×108]] / [0.57×10-6]
= [1.9878×10-25] / [0.57×10-6]
= 3.487×10-19 J
Rate of Emission of Quanta, R = P / E
Where
P = Power of the bulb
E = Energy of photon
R = [25] / [3.487×10-19]
= 7.169×1019 s-1
Therefore, the rate of emission of quanta is 7.17×1019 s-1.