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(B) Given: Power of the source $P = 60 \,W$.Wavelength of light $\lambda = 662.5 \,nm = 6.625 	imes 10^{-7} \,m$.The energy of a single photon is giv

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$2 	imes 10^{20}$

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(B) Given: Power of the source $P = 60 \,W$.Wavelength of light $\lambda = 662.5 \,nm = 6.625 	imes 10^{-7} \,m$.The energy of a single photon is given by $E = \frac{hc}{\lambda}$.Substituting the values ($h = 6.625 	imes 10^{-34} \,J \cdot s$ and $c = 3 	imes 10^8 \,m/s$):$E = \frac{6.625 	imes 10^{-34} 	imes 3 	imes 10^8}{6.625 	imes 10^{-7}} = 3 	imes 10^{-19} \,J$.The number of photons emitted per second $(n)$ is given by the ratio of total power to the energy of one photon:$n = \frac{P}{E} = \frac{60}{3 	imes 10^{-19}} = 20 	imes 10^{19} = 2 	imes 10^{20}$ photons/second.

$A$ $60 \,W$ source emits monochromatic light of wavelength $662.5 \,nm$. The number of photons emitted per second is

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