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(C) The power $P$ of the transmitter is the total energy emitted per second. Energy of one photon is given by $E = \frac{hc}{\lambda}$. The number of

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$3 	imes 10^{20}$

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(C) The power $P$ of the transmitter is the total energy emitted per second. Energy of one photon is given by $E = \frac{hc}{\lambda}$. The number of photons $n$ emitted per second is given by $n = \frac{P}{E} = \frac{P\lambda}{hc}$. Given: $P = 100 \ W$,$\lambda = 540 	imes 10^{-9} \ m$,$h = 6 	imes 10^{-34} \ J \cdot s$,$c = 3 	imes 10^8 \ m/s$. Substituting the values: $n = \frac{100 	imes 540 	imes 10^{-9}}{6 	imes 10^{-34} 	imes 3 	imes 10^8}$. $n = \frac{54000 	imes 10^{-9}}{18 	imes 10^{-26}} = \frac{5.4 	imes 10^4 	imes 10^{-9}}{1.8 	imes 10^{-25}} = 3 	imes 10^{20}$. Thus,the number of photons emitted per second is $3 	imes 10^{20}$.

$A$ transmitter of $100 \ W$ power emits electromagnetic waves of wavelength $540 \ nm$. How many photons are emitted per second? (Given: $h = 6 \times 10^{-34} \ J \cdot s$,$c = 3 \times 10^8 \ m/s$)

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