Year 12 Cambridge Physics: Classical Mechanics v Light Particle Collision

[johnconnor]

Question:
A stream of particles, each of mass m and having kinetic energy E, is collimated into a parallel beam of cross-sectional area A. The particles are incident normally on a smooth plane surface at rate n and they rebound elastically. Derive an expression for the pressure on the surface in terms of A, m, E and n. Explain why the pressure would be different if the surface were rough so that the particles rebounded at various angles.

An electric light bulb emits 20W of radiation uniformly in all directions. What is the maximum radiation pressure on a surface placed 2.0m away from the bulb? State the conditions under which this will occur?

Attempt:
E_k = 0.5mu^2 \Rightarrow u=\sqrt{2E/m}
\text{Ft} = 2m\sqrt{2E/m}
F = Ftn
P= \dfrac{2mn\sqrt{2E/m}}{A}
P = \dfrac{2n\sqrt{2EM}}{A}

2nd part: incident momentum is added a factor of sin/cosine and therefore effective impulse decreases (blah blah blah..).

I don't know how to start the light bulb part. Does λ=h/p come into use? The wavelength of light covers a range of values; which should I choose? Can anyone please give me some hints? Thank you!

 

[Infinitum]

Your first part seems correct to me. Apply the same principle for the second, too. The force exerted by light would be the momentum transferred per unit second to the object. Pressure, hence would be the momentum transferred per unit second per unit area. So,

Pressure = \frac{\Delta p}{At}

But the intensity is also the energy density. So, we have

P = \frac{I}{c}

where c is the speed of the photons. I think you should be able to do it from here

 

[johnconnor]

But the intensity is also the energy density. So, we have

P = \frac{I}{c}

where c is the speed of the photons. I think you should be able to do it from here

Ah. I didn't know about this formula. Maybe I've forgotten it. Thanks! Will work on it!

 

[Infinitum]

Ah. I didn't know about this formula. Maybe I've forgotten it. Thanks! Will work on it!

Oh, it is easily derivable

Did you notice I derived this for an absorbing surface?? Can you see what it would be for a reflecting surface?

 

[johnconnor]

Your first part seems correct to me. Apply the same principle for the second, too. The force exerted by light would be the momentum transferred per unit second to the object. Pressure, hence would be the momentum transferred per unit second per unit area. So,

Pressure = \frac{\Delta p}{At}

But the intensity is also the energy density. So, we have

P = \frac{I}{c}

where c is the speed of the photons. I think you should be able to do it from here

So I reattempted the final part in lieu of Infinitum's tip, and my working is as such:

Consider a beam of light with energy E incident on a flat surface area A 2.0m away from the light bulb.

E=hf; Power=E/t; a^2=A

Power = E/t

Power = hf/t

Since t = 2/c,

Power = hfc/2

Force x 2 x c/2 = hfc/2

Force/A = hf/2

Ah i just bullgarbageed. Sorry.
There are plenty of errors in my attempt: I didn't try to find the value of A, and I'm assuming that intensity remains the same after traveling 2metres (which contradicts the inverse square law). But what other mistakes have I made? What assumptions which I thought to be applicable but fallacious in this context?

Also, could you please elaborate on your post? Thank you...

 

[ehild]

So I reattempted the final part in lieu of Infinitum's tip, and my working is as such:

Consider a beam of light with energy E incident on a flat surface area A 2.0m away from the light bulb .

E=hf; Power=E/t; a^2=A

Power = [STRIKE]E/t[/STRIKE]

Power = [STRIKE]hf/t[/STRIKE]

Since t = 2/c,

Power = [STRIKE]hfc/2[/STRIKE]

E=hf is the energy of a single photon. The power of the source is the energy emitted in unit time. The power of the source has nothing to do with the time the photons reach from the source to the surface.
You get the number of photons N emitted in one second if you divide the power of the source by the energy of a single photon. These photons are emitted in every directions. You need to find the number of photons n falling on the area A at 2 m distance in unit time. How do you it?

To get the pressure you need to do the same as before: Find the change of the momentum Δp of a single photon upon collision. As n photon arrives in unit time the force they exert on the surface is F=nΔp. Divide it by the surface area to get the pressure.

Now the main thing: the momentum of a photon. It depends on the wavelength: p=h/λ, or p=hf/c in terms of the frequency.
(http://en.wikipedia.org/wiki/Photon) You certainly have learned it together with the formula for the energy of a photon E=hf.

ehild

 

[Infinitum]

Ehild pretty much said it all

You can also take a look at radiation pressure