Question
A strip of metal foil with a mass of 5.00×107 kg5.00\times 10^{-7}\textrm{ kg} suspended in a vacuum and exposed to a pulse of light. The velocity of the foil changes from zero to 1.00×103 m/s1.00\times 10^{-3}\textrm{ m/s} in the same direction as the initial light pulse, and the light pulse is entirely reflected from the surface of the foil. Given that the wavelength of the light is 450 nm, and assuming that this wavelength is the same before and after the collision, how many photons in the pulse collide with the foil?
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Final Answer

1.70×1017 photons1.70\times 10^{17}\textrm{ photons}

Solution video

OpenStax College Physics for AP® Courses, Chapter 29, Problem 8 (Test Prep for AP® Courses)

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Video Transcript
This is College Physics Answers with Shaun Dychko. A metal foil with a mass of 5.00 times 10 to the minus 7 kilograms is suspended in a vacuum and it's exposed to a pulse of light and the foil has a velocity that changes from zero initially to 1.00 times 10 to the minus 3 meters per second in the same direction as the initial light pulse so if the light was going this way, the foil will eventually have a velocity in that same direction and this light pulse is reflected from the foil and the wavelength of the light doesn't change. Now this pulse of light will have some momentum due to all the photons that are in this pulse and the final momentum of the light will be the negative of whatever momentum it had initially because it's now going in the opposite direction and since the wavelength hasn't changed, its momentum will not change either since momentum is Planck's constant divided by the wavelength for a single photon. So the wavelength is 450 nanometers and the change in momentum is the final momentum minus the initial momentum and the final momentum is negative momentum of a photon minus the initial momentum, which is momentum of a photon this works out to negative 2 times the momentum of a single photon. Now the change in momentum of the foil— this f here represents 'foil,' it's not 'final'... sorry if that's a bit confusing— it equals the opposite to the total change in momentum of all the photons so there are N—number of photons in this pulse— so if we have this ΔP representing the change in momentum of a single photon this N represents the change in all the photons put together and the change has to be opposite to the change in the foil since momentum has to be conserved— momentum can't be created nor destroyed— so whatever change there is in the momentum of the foil has to be compensated for in the opposite direction to the change in momentum of the photons. Okay! So the change in momentum of the foil is mass times final velocity minus mass times initial velocity but the initial velocity is zero so we didn't bother with that term so we just have < i>mv f is the change in momentum of the foil— f here is 'final'... there we go— and on the right hand side, we have negative N times negative 2 times the momentum of a single photon this negative and this negative both make positive so we have 2N momentum of a single photon. Momentum of a single photon is Planck's constant over wavelength and we can substitute that in for P γ and we do that here and then we solve for N by multiplying both sides by λ over 2 times Planck's constant. So the number of photons then is the mass of the foil times its final velocity times the wavelength of the light divided by 2 times Planck's constant and that works out to 1.70 times 10 to the 17 photons.