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A 200 W sodium street lamp emits yellow light of wavelength 0.6 $\mu$m. Assuming it to be 25% efficient in converting electrical energy to light, the number of photons of yellow light it emits per second is
A
$3\times10^{19}$
B
$1.5\times10^{20}$
C
$6\times10^{18}$
D
$62\times10^{20}$
Detailed Solution
Light energy emitted per second $= 200\times\frac{25}{100} = 50$ W = 50 J/s
Energy of one photon $= \frac{hc}{\lambda} = \frac{6.6\times10^{-34}\times3\times10^8}{0.6\times10^{-6}} = 3.3\times10^{-19}$ J
Number of photons per second $= \frac{50}{3.3\times10^{-19}} = \frac{50\times0.6\times10^{-6}}{6.6\times10^{-34}\times3\times10^8} = 1.5\times10^{20}$
Energy of one photon $= \frac{hc}{\lambda} = \frac{6.6\times10^{-34}\times3\times10^8}{0.6\times10^{-6}} = 3.3\times10^{-19}$ J
Number of photons per second $= \frac{50}{3.3\times10^{-19}} = \frac{50\times0.6\times10^{-6}}{6.6\times10^{-34}\times3\times10^8} = 1.5\times10^{20}$