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(B) The energy of a single photon is given by $E = \frac{hc}{\lambda}$.Total power emitted as light is $P = nE$,where $n$ is the number of photons emi

Vedclass · Accepted Answer

$27.17 	imes 10^{18}$

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(B) The energy of a single photon is given by $E = \frac{hc}{\lambda}$.Total power emitted as light is $P = nE$,where $n$ is the number of photons emitted per second.Thus,$P = \frac{nhc}{\lambda}$,which implies $n = \frac{P \lambda}{hc}$.The useful power $P$ is $8 \%$ of $150 	ext{ W}$,so $P = \frac{8}{100} 	imes 150 = 12 	ext{ W}$.Using $h = 6.63 	imes 10^{-34} 	ext{ J s}$,$c = 3 	imes 10^8 	ext{ m/s}$,and $\lambda = 4500 	imes 10^{-10} 	ext{ m}$:$n = \frac{12 	imes 4500 	imes 10^{-10}}{6.63 	imes 10^{-34} 	imes 3 	imes 10^8}$$n = \frac{54000 	imes 10^{-10}}{19.89 	imes 10^{-26}}$$n \approx 27.15 	imes 10^{18} \approx 27.17 	imes 10^{18}$ photons per second.

$A$ bulb lamp emits light of mean wavelength of $4500 \mathring A$. The lamp is rated at $150 \text{ W}$ and $8 \%$ of the energy appears as emitted light. How many photons are emitted by the lamp per second?

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