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(A) Total energy emitted in $0.1 \, s$ is given by $E = P 	imes t = 10^{-3} \, W 	imes 0.1 \, s = 10^{-4} \, J$.The energy of a single photon is $E_

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$50$

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(A) Total energy emitted in $0.1 \, s$ is given by $E = P 	imes t = 10^{-3} \, W 	imes 0.1 \, s = 10^{-4} \, J$.The energy of a single photon is $E_{photon} = \frac{hc}{\lambda} = \frac{6.63 	imes 10^{-34} \, J \, s 	imes 3.00 	imes 10^{8} \, m \, s^{-1}}{1000 	imes 10^{-9} \, m} = 1.989 	imes 10^{-19} \, J$.The number of photons $n$ is given by $n = \frac{E}{E_{photon}} = \frac{10^{-4} \, J}{1.989 	imes 10^{-19} \, J} \approx 5.027 	imes 10^{14}$.Expressing this as $x 	imes 10^{13}$,we get $n = 50.27 	imes 10^{13}$.Rounding to the nearest integer,$x = 50$.

The number of photons emitted by a monochromatic (single frequency) infrared range finder of power $1 \, mW$ and wavelength of $1000 \, nm$,in $0.1 \, s$ is $x \times 10^{13}$. The value of $x$ is ..... .
(Nearest integer)
$(h=6.63 \times 10^{-34} \, J \, s, c=3.00 \times 10^{8} \, m \, s^{-1})$

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The number of photons emitted by a monochromatic (single frequency) infrared range finder of power $1 \, mW$ and wavelength of $1000 \, nm$,in $0.1 \, s$ is $x \times 10^{13}$. The value of $x$ is ..... .(Nearest integer)$(h=6.63 \times 10^{-34} \, J \, s, c=3.00 \times 10^{8} \, m \, s^{-1})$