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(C) The energy of a single photon is given by $E = \frac{hc}{\lambda}$.Substituting the values: $E = \frac{6.6 	imes 10^{-34} 	imes 3 	imes 10^8}{6

Vedclass · Accepted Answer

$2.5 	imes 10^{13}$

Vedclass · Answer

(C) The energy of a single photon is given by $E = \frac{hc}{\lambda}$.Substituting the values: $E = \frac{6.6 	imes 10^{-34} 	imes 3 	imes 10^8}{660 	imes 10^{-9}} = 3 	imes 10^{-19} J$.The total energy emitted by the lamp in time $t = 100 ms = 0.1 s$ is $E_{total} = P 	imes t = 300 	imes 0.1 = 30 J$.The intensity at a distance $r = 10 m$ is distributed over a sphere of area $A_{sphere} = 4\pi r^2 = 4\pi(10)^2 = 400\pi m^2$.The area of the sensor opening with radius $r_s = 1 cm = 10^{-2} m$ is $A_{sensor} = \pi r_s^2 = \pi(10^{-2})^2 = \pi 	imes 10^{-4} m^2$.The energy incident on the sensor is $E_{sensor} = E_{total} 	imes \frac{A_{sensor}}{A_{sphere}} = 30 	imes \frac{\pi 	imes 10^{-4}}{400\pi} = \frac{30 	imes 10^{-4}}{400} = 0.075 	imes 10^{-4} = 7.5 	imes 10^{-6} J$.The number of photons $n$ is $\frac{E_{sensor}}{E} = \frac{7.5 	imes 10^{-6}}{3 	imes 10^{-19}} = 2.5 	imes 10^{13}$.

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