Question

A 25 watt bulb emits monochromatic yellow light of wavelength of 0.57µm. Calculate the rate of emission of quanta per second.

Answer 1

Power of bulb, P= 25 Watt = 25 Js
Energy of one photon, E=$\frac{hc}{\lambda}$
Substituting the values in the given expression of E: E = $\frac{(6.626\times10^{-34})(3\times10^8)}{0.57\times 19^{-6}}$ = 34.87 × 10-20 J
Rate of emission of quanta per second =$\frac{25}{34.87}$ × 10 -20 = 7.169 × 1019 s -1.