A $60 \mathrm{~W}$ bulb emits monochromatic light of wavelength $663 \mathrm{~nm}$. The exact number

Question

A $60 \mathrm{~W}$ bulb emits monochromatic light of wavelength $663 \mathrm{~nm}$. The exact number of photons emitted per second by the bulb is (Given $h = 6.63 	imes 10^{-34} \mathrm{~Js}$ and $c = 3 	imes 10^8 \mathrm{~m/s}$)

Professor PC Thomas Classes · Answer

(a) : The energy of a single photon is $E = \frac{hc}{\lambda} = \frac{6.63 	imes 10^{-34} 	imes 3 	imes 10^8}{663 	imes 10^{-9}} = \frac{19.89 	imes 10^{-26}}{6.63 	imes 10^{-7}} = 3 	imes 10^{-19} \mathrm{~J}$. The power of the bulb is $P = 60 \mathrm{~W} = 60 \mathrm{~J/s}$. The number of photons emitted per second is $N = \frac{P}{E} = \frac{60}{3 	imes 10^{-19}} = 20 	imes 10^{19} = 2 	imes 10^{20}$ photons per second.