Light with an energy flux of 9 Wcm^{-2} fall | vedclass

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Question

(C) The energy flux $I = 9 \ Wcm^{-2} = 9 	imes 10^4 \ Wm^{-2}$.The area $A = 20 \ cm^2 = 20 	imes 10^{-4} \ m^2$.The time $t = 1 \ hour = 3600 \ s$

Vedclass · Accepted Answer

$2.16 	imes 10^{-3} \ kgms^{-1}$

Vedclass · Answer

(C) The energy flux $I = 9 \ Wcm^{-2} = 9 	imes 10^4 \ Wm^{-2}$.The area $A = 20 \ cm^2 = 20 	imes 10^{-4} \ m^2$.The time $t = 1 \ hour = 3600 \ s$.The speed of light $c = 3 	imes 10^8 \ ms^{-1}$.For a non-reflecting surface, the momentum $p$ delivered by radiation is given by $p = \frac{U}{c} = \frac{IAt}{c}$.Substituting the values:$p = \frac{(9 	imes 10^4) 	imes (20 	imes 10^{-4}) 	imes 3600}{3 	imes 10^8}$.$p = \frac{180 	imes 3600}{3 	imes 10^8} = \frac{648000}{3 	imes 10^8} = 216000 	imes 10^{-8} \ kgms^{-1}$.$p = 2.16 	imes 10^{-3} \ kgms^{-1}$.

Light with an energy flux of $9 \ Wcm^{-2}$ falls on a non-reflecting surface at normal incidence. If the surface has an area of $20 \ cm^2$, the total momentum delivered for complete absorption in one hour is:

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