The force on a hemisphere of radius 1, cm if | vedclass

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(C) The momentum of each photon is $p = \frac{h}{\lambda} = \frac{6.63 	imes 10^{-34}}{500 	imes 10^{-9}} = 1.33 	imes 10^{-27} \, kg \cdot m/s$.Th

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$5.22	imes10^{-9}\, N$

Vedclass · Answer

(C) The momentum of each photon is $p = \frac{h}{\lambda} = \frac{6.63 	imes 10^{-34}}{500 	imes 10^{-9}} = 1.33 	imes 10^{-27} \, kg \cdot m/s$.The number of photons striking per unit area per unit time is $n = \frac{I}{hc/\lambda} = \frac{I \lambda}{hc}$.Given $I = 0.5 \, W/cm^2 = 0.5 	imes 10^4 \, W/m^2$,the number of photons per unit area per second is $n = \frac{0.5 	imes 500 	imes 10^{-9}}{6.63 	imes 10^{-34} 	imes 3 	imes 10^8} \approx 1.25 	imes 10^{22} \, photons/(m^2 \cdot s)$.For perfectly inelastic collisions,the force exerted by a photon is $F = \frac{dp}{dt} = p \cdot n \cdot A_{eff}$.The effective area $A_{eff}$ projected perpendicular to the beam is the area of the circular base of the hemisphere,$A = \pi r^2 = \pi 	imes (0.01)^2 = \pi 	imes 10^{-4} \, m^2$.Force $F = (1.33 	imes 10^{-27}) 	imes (1.25 	imes 10^{22}) 	imes (\pi 	imes 10^{-4}) \approx 5.22 	imes 10^{-9} \, N$.

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