Pressure Exerted by Electron Question

[joey7]

Appreciate if someone could give me a hand with this question has me a bit lost;

A 500W lamp directs a beam of yellow light, of wavelength 580nm, into a perfect reflecting surface of area 4.0cm^2.

"An electron beam is directed onto reflecting surface. If electrons in the beam each have the same energy as a photon of yellow light" (initially forgot this bit) determine the number of electrons that are incident in order to produce a pressure of 0.533 Nm^-2

[mark_alec]

Electrons? Are you sure you don't mean photons?

[Hancock]

So, we know that:




The 10^-4 comes from the fact that we need the units of Area to be in m^2.

ASSUMING HE MEANT PHOTONS.

From De Broglie's wavelength relation:


For one photon:


So by ratios, we know that 1 photon produces 1.1431e-27 kgm/s.
And we want x photons to produce 0.0002132 kgm/s.

Therefore:
X =

[Alwin]

Electrons? Are you sure you don't mean photons?

iirc, this was a q from checkpoints joey7?

if it was, then yes mark_alec it was photons.

hint: work out the momentum of the photons, then the force they apply

EDIT: look at Hancock, beat me to it while i was looking around for my old checkpoints book to double check
Electrons do exhibit wave-like properties.. but never seen a lamp that beams out electrons of yellow colour XD

[joey7]

Hey guys sorry, summarised question and left out the part that said "An electron beam is directed onto reflecting surface, if the electrons in the beam each have the same energy as a photon of yellow light then..."

Thanks for the explanations what I dont understand though is: can you equate momentum and force? momentum is Newtons per second and force is just newtons?

[Hancock]

Look above. You cannot equate momentum and force because they are completely separate quantities.
F = dp/dt for those who are mathematical inclined in Year 12.

[SocialRhubarb]

If the surface is perfectly reflecting, don't you impart double the momentum of the photon onto the surface? Because you aren't just stopping the photon, you're giving it an equal momentum in the opposite direction, and so the force on the surface doubles to double the change in momentum.

Also, momentum isn't Newtons per second, it's Newtons times seconds. So it's actually the other way round. Force is momentum per second. What Hancock did was he found the momentum of one photon, and he found the number of photons needed per second to exert said force.

I don't really like the question, to be honest, because the number of photons you're emitting per second is already determined by the power of the lamp, which you apparently don't use at all, by the way.

[joey7]

If the surface is perfectly reflecting, don't you impart double the momentum of the photon onto the surface? Because you aren't just stopping the photon, you're giving it an equal momentum in the opposite direction, and so the force on the surface doubles to double the change in momentum.

Also, momentum isn't Newtons per second, it's Newtons times seconds. So it's actually the other way round. Force is momentum per second. What Hancock did was he found the momentum of one photon, and he found the number of photons needed per second to exert said force.

I don't really like the question, to be honest, because the number of photons you're emitting per second is already determined by the power of the lamp, which you apparently don't use at all, by the way.

Oh right so we are finding out the force per second to produce this force that's what was confusing me, Nar the question makes sense I just accidentally left a pretty important bit out I edited the question and put it in but even with above explanation I understand how to go about getting the answer

Also I don't know why the momentum is not doubled

[SocialRhubarb]

Actually, we're finding the change in momentum per second, which is force.

[joey7]

Oh right so another way to look at it is that the seconds will cancel out and just leave you with newtons (force)?