A bulb has a power of 60 , mW and emits light | vedclass

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Question

$(A)$ The power $P$ emitted by the bulb is given by the total energy of $n$ photons emitted per second: $P = n \cdot E = n \cdot \frac{hc}{\lambda}$.R

Vedclass · Accepted Answer

$1.8 	imes 10^{17} \, 	ext{photons/second}$

Vedclass · Answer

$(A)$ The power $P$ emitted by the bulb is given by the total energy of $n$ photons emitted per second: $P = n \cdot E = n \cdot \frac{hc}{\lambda}$.Rearranging for $n$, we get $n = \frac{P \lambda}{hc}$.Given values: $P = 60 \, mW = 60 	imes 10^{-3} \, W$, $\lambda = 6000 \, \mathring{A} = 6000 	imes 10^{-10} \, m$, $h = 6.62 	imes 10^{-34} \, J \cdot s$, and $c = 3 	imes 10^8 \, m/s$.Substituting these values:$n = \frac{(60 	imes 10^{-3}) 	imes (6000 	imes 10^{-10})}{(6.62 	imes 10^{-34}) 	imes (3 	imes 10^8)}$$n = \frac{360000 	imes 10^{-13}}{19.86 	imes 10^{-26}}$$n \approx 1.812 	imes 10^{17} \approx 1.8 	imes 10^{17} \, 	ext{photons/second}$.

$A$ bulb has a power of $60 \, mW$ and emits light of wavelength $6000 \, \mathring{A}$. Calculate the number of photons emitted by the bulb per second.

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