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Question

(D) The intensity $I$ at a distance $r = 5 \ m$ from the source is given by $I = \frac{P}{4 \pi r^2}$.Substituting the values: $I = \frac{942}{4 	ime

Vedclass · Accepted Answer

$4 	imes 10^{19}$

Vedclass · Answer

(D) The intensity $I$ at a distance $r = 5 \ m$ from the source is given by $I = \frac{P}{4 \pi r^2}$.Substituting the values: $I = \frac{942}{4 	imes 3.14 	imes 5^2} = \frac{942}{12.56 	imes 25} = \frac{942}{314} = 3 \ W/m^2$.The energy of a single photon is $E = \frac{hc}{\lambda} = \frac{6.6 	imes 10^{-34} 	imes 3 	imes 10^8}{660 	imes 10^{-9}} = \frac{19.8 	imes 10^{-26}}{6.6 	imes 10^{-7}} = 3 	imes 10^{-19} \ J$.The photon flux $\phi$ is the number of photons per unit area per unit time,given by $\phi = \frac{I}{E}$.$\phi = \frac{3}{3 	imes 10^{-19}} = 1 	imes 10^{19} \ \frac{	ext{photon}}{m^2 \cdot s}$.

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