A 200, W sodium street lamp emits yellow ligh | vedclass

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(A) The total electrical power input is $P_{in} = 200\, W$.Given the efficiency is $25\%$, the effective power output as light is $P_{out} = 0.25 	im

Vedclass · Accepted Answer

$1.5 	imes 10^{20}$

Vedclass · Answer

(A) The total electrical power input is $P_{in} = 200\, W$.Given the efficiency is $25\%$, the effective power output as light is $P_{out} = 0.25 	imes 200\, W = 50\, W$.The energy of a single photon is given by $E = \frac{hc}{\lambda}$.If $n$ is the number of photons emitted per second, then the total power is $P_{out} = nE = n \frac{hc}{\lambda}$.Rearranging for $n$, we get $n = \frac{P_{out} \lambda}{hc}$.Substituting the values: $n = \frac{50 	imes 0.6 	imes 10^{-6}}{6.63 	imes 10^{-34} 	imes 3 	imes 10^8}$.$n = \frac{30 	imes 10^{-6}}{19.89 	imes 10^{-26}} \approx 1.5 	imes 10^{20}$ photons per second.

$A$ $200\, W$ 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:

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