# Quantum mechanics - Compton effect

1. May 14, 2014

### Feodalherren

1. The problem statement, all variables and given/known data

A photon with energy 3.98E-19 J hits a stationary electron. The photon is deflected 90 degrees and the electron recoils back and down. At what angle is the electron deflected?
Hint: the electron will be going at << .10c

2. Relevant equations

3. The attempt at a solution
Skipping some simple plug and chug steps this is what I found:

λ(photon) = 499nm
P(photon) and therefore also Pi of the system = 1.328E-27 kgm/s.

This is where I get very confused.

Since the photon is deflected 90 degrees the Compton effect is negligible:

$\Delta \lambda = \frac{h}{(9.11E-31kg)c} = 2.426E-12m$.

So there is essentially no change in lambda. Therefore the energy of the photon is internally conserved. This means that the energy of the photon is essentially the same before and after it hits the photon. If the energy is conserved before and after, how on earth can the electron gain any kinetic energy!? This seems to violate the conservation of energy.

Yet when I do the momentum part I get

$P_{photon}=1.328E-27kgm/s = P_{Xe-}=(9.11E-31kg)(V_{Xe-})$

$V_{Xe-}=V_{Ye-}=1458kgm/s$ in the positive X and negative Y direction.

Therefore the angle is 45 degrees.

But as I said. Out of an energy point of view I can't make sense of it.

2. May 14, 2014

### Simon Bridge

You are told that the final speed of the electron is <0.1c ... what is special about this?
Note: just because the photon lost a very small proportion of it's energy does not mean that it can be neglected.

Aside:
Best practice: Do everything in electron-volts.
units are: eV for energy, eV/c for momentum, and eV/c^2 for mass.

So the photon energy is 2.48eV and it's initial momentum is 2.48eV/c
electron rest-mass is 0.511eV/c^2 and so on.

If you also write all speeds as a fraction of the speed of light a lot of equations get simple.

3. May 14, 2014

### haruspex

There is a difference between the energy being essentially the same and its being exactly the same.
According to my calculation, the photon loses a mere 0.0005% of its energy, but that's enough to send the electron off at 2km/s.

4. May 15, 2014

### Feodalherren

Since the electron is going at less than 10 % of the SOL I can ignore relativity for the electron's momentum and energy.

I don't understand how I'm supposed to make the amount of energy lost count in anywhere. The Compton effect is so small that my sig figs knock it out of the picture. At that point I have nowhere to go with energy....!?

If Ei=2.48 eV then the final energy of the photon is:

2.4846eV=2.48eV

Okay, how does that even help me? Where do I even need to use conservation of energy? I can do the whole thing without even concerning myself with it yet the solution states that we must use both principles.

5. May 15, 2014

### Simon Bridge

Sounds like you need a different way to handle the problem besides the compton effect equation.
How would you normally do conservation of energy and momentum problems?

6. May 15, 2014

### Feodalherren

I would calculate the initials or finals, depending on what's given set them equal to each other and see what I can solve for.

So in this problem I'm given Ei and I can easily solve for Pi. So that's basically what I did. But I ended up ignoring energy. I never actually used the conservation of energy which is what is confusing me. Where and why am I supposed to be using it?

I remember a similar problem from classical mechanics with billiard balls where the energy was conserved in a case that couldn't happen because momentum wasn't conserved in both cases. If memory serves me right one of the cases was where the balls stuck together and the other was where the kinetic energy was transferred. This feels similar but different.

7. May 15, 2014

### haruspex

Maybe I'm missing something, but it seems to me this is just like solving a billiard ball problem with elastic collisions. You will need both conservation of momentum and conservation of energy to solve it. The only difference is the way you compute the momenta of the photon - i.e. you use the Compton effect formula.

8. May 15, 2014

### Simon Bridge

You didn't really - you have three unknowns after all.

If you use the Compton effect equation to get the final energy and momentum of the photon, you don't need it. It's already accounted for. The energy lost by the photon is gained by the electron.
http://en.wikipedia.org/wiki/Compton_scattering#Derivation_of_the_scattering_formula

Well this is just like that, only energy and momentum are both conserved.
(the balls sticking together means the collision is not elastic.)

What happens if you just naively go through the conservation of energy and momentum without worrying about the Compton effect equation?

9. May 16, 2014

### Feodalherren

Ah in that case yes I can use both conservational principles - I end up with the same answer.

10. May 16, 2014

### Simon Bridge

Does this approach satisfy your concerns re conservation of energy?

11. May 16, 2014

### Feodalherren

Same answer as I got using the previous method - the angle is 45 degrees. I suppose my confusion came from the fact that I thought that the photon would have to lose significantly more energy to give the electron the big velocity. It seems that an almost infinitesimal amount of energy loss from the photon will shoot off the electron at fast speeds.

12. May 16, 2014

### Simon Bridge

Technically the angle will be slightly less than 45deg ... but too close to call.
But you can work it out for hockey-pucks ... incoming puck is deflected through 90deg, but has almost the same speed, what happens to the other puck? Is energy conserved? How about if the second puck is very small compared with the first one?

Note: the photon did not give the electron a big velocity - so it dd not have to lose a lot of it's energy.
"Big" is a relative term ... part of the whole point in the problem is that the electron's speed was very small compared with the speed of light.

The energy loss was not infinitesimal nor negligible though - it was just smaller than the uncertainty implied in the numbers (so it vanishes in the sig figs). Just because you didn't measure it don't mean it ain't there.