Jump to content

Compton effect question


mooeypoo

Recommended Posts

I started taking QM this semester and I am a bit confused with this question:

 

"In the Compton effect, if the incident radiation has wavelength [math]\lambda_{0}[/math], what is the maxima wavelength that the scattered radiation can have, and for what angle will this happen?"

 

So, maxima/minima is calculated using a tangent, so I figured I will differentiate the equation:

 

[math]\lambda_{0} - \lambda = \frac{h}{mc} ( 1-\cos{\theta} )[/math]

 

With respect to [math]\lambda[/math], but that made no sense (=0?).

 

So, I figured I'll go a step back on the derivation of the equation and try from there:

 

[math]\frac{h^2}{\lambda_{0}\lambda}(1-\cos{\theta})=mc(\frac{h}{\lambda_{0}}-\frac{h}{\lambda})[/math]

 

But that doesn't come out well either (0=0)

 

So I resorted to looking at the graph I was given, seeing that [math]\lambda_{1}-\lambda_{0}=\frac{2\pi}{mc}[/math] when [math]\theta=\pi[/math] (which is the max).

 

But that strikes me as a bit of a cop-out. Is there a way to calculate this question algebraically? What did I miss?

 

The next part of the question makes me think I am way off track here. It asks "Can a free electron absorb a photon while conserving energy and momentum? (Hint: think of part (a): what wavelength must the 'emitted' photon have if it has zero energy?)"

 

Err the only connection i can think of is E=hv or [math]E=\frac{hc}{\lambda}[/math] but then if E=0, then lambda should be ... like.. infinity.

 

Help!

Edited by mooeypoo
Link to comment
Share on other sites

I thought about that in the beginning, but

[math]\frac{d\lambda}{d\theta}=\frac{h}{mc}\sin{\theta}=0[/math]

....

 

Okay, as I was typing it again, I answered my own following question. Theta=pi, and

[math]

\lambda_{1}-\lambda_{0}=\frac{2\pi}{mc}

[/math] just like the graph says. Okay.

 

Now for part b, though, I'm not sure why I even need to consider a case where E=0, but when I do, then [math]E=0=hc/\lambda[/math] and lambda is infinite.

 

What does that have to do with free electron absorbing a photon while conserving energy and momentum? I am not sure I understand the connection between the question and the hint and the first part.

Link to comment
Share on other sites

Now for part b, though, I'm not sure why I even need to consider a case where E=0, but when I do, then [math]E=0=hc/\lambda[/math] and lambda is infinite.

 

What does that have to do with free electron absorbing a photon while conserving energy and momentum? I am not sure I understand the connection between the question and the hint and the first part.

 

If lambda is infinite, is there a solution to the problem?

 

You can solve it directly as well, using conservation of momentum and energy (it's relativistic)

Link to comment
Share on other sites

If lambda is infinite, is there a solution to the problem?

 

You can solve it directly as well, using conservation of momentum and energy (it's relativistic)

No, Lambda can't be infinite, of course not, I just don't see why I'm measuring lambda -- the question asks if the electron can absorb a photon while conserving energy and momentum; Why do I equate the photon's energy to zero?

 

When the energy is zero, the lambda is infinite, which makes no sense. I think that the energy is set to zero to "imagine" a situation where the photon's entire energy passed to the electron, and now the photon has zero energy, and I am supposed to conclude that this can't be. BUT I don't quite understand -- a photon is made of energy, it has no mass, if it is absorbed in an electron, why do I even need to consider a photon that has no energy?? The photon 'vanished', there's no "resulting" photon after the collision, there's just an electron, because the hypothetical situation is that the electron took *all* of the photon's energy and momentum.

Why can't the electron just get all the energy, in which case the equation of E=0 is meaningless....

 

The only reason I can think of why this can't happen is because the electron *does* have a mass, in which case it cannot move at the speed of light, which means it couldn't have taken the total energy and the total momentum from the phton, but (a) I'm not sure how to show it and (b) I'm not sure what this has to do with E=0.

 

I am hoping I manage to explain myself clearly :\ it's a bit confusing.

 

~moo


Merged post follows:

Consecutive posts merged

Err, I just made my own case <sob>

 

Part © asks to give another derivation of the answer in part (b) by considering the photon absorption process in a reference frame in which the electron after the absorption of the photon is at rest and 'play the movie backwards'. Does anything go wrong?

 

So, in this case, I can see why there would be a problem (I need to work on the actual solution still, but it makes more sense) because the electron would have to be moving at the speed of light, and that's not possible to a particle with mass..... but why the E=0 situation?

Link to comment
Share on other sites

If the photon is completely absorbed, that's the same as saying the scattered photon has no energy. But you can see from the equation and part a that there is, in fact, a maximum wavelength of the scattered photon, which is finite — so not scattering a photon is not an option.

Link to comment
Share on other sites

swansont, why does the reaction has to conclude with a scattered photon? a scattered photon assumes in advance that the photon wasn't completely absorbed, no? If the photon *was* absorbed, then there's no scattered photon, which can also explain why the lambda is infinite (impossible, so it's non existing).

 

I just have this feeling that I am going in the wrong direction here; I am trying to explain why something is assumed by assuming something that's assumed.. it makes no sense to me. If the collision was inellastic, the photon's energy (which is to say the photon itself) was absorbed completely in the electron, and there is no resulting photon. Why are we not assuming that to begin with?

 

I do understand that an electron cannot travel at the speed of light, which means that if the photon was completely absorbed, there must be some excess energy that remains, and hence the electron cannot absorb the entire photon, but that can be explained with special relativity and 'simple' collisions..

 

It seems a bit weird to me that we are already assuming that there *must* be an emitted photon, but seeing as the lambda of that emit photon makes no sense, then instead of concluding that the emitted photon can't exist, we conclude that the absorption can't exist.

 

Isn't that backwards? It's assuming on the assumption of the assumption. It's weird :\

Link to comment
Share on other sites

I've had a long chat with ajb and timo on the IRC chan about this.

 

What we did was equate the initial energy and initial momentum to the final energy and final momentum, and I saw that they create a nonsensical equation. That really hammered things home:

 

Assuming 100% transfer of momentum:

[math]p_{photon}+0=0+p_{e final}[/math]

[math]p_{e}=p_{photon}[/math]

 

Assuming 100% transfer of energy:

[math]pc+mc^2=0+E_{e}[/math]

[math]E_{e}=pc+mc^2[/math]

 

And the equation of energy for the final photon should be:

[math]E^2=m^2c^4+p^2c^2[/math]

 

So, combining the above to see the final energy of the electron:

 

[math](pc+mc^2)^2=m^2c^4+p^2c^2[/math]

[math]p^2c^2+m^2c^4+2pcmc^2=m^2c^4+p^2c^2[/math]

[math]2pcmc^2=0[/math]

And we know that [math]m \neq 0[/math] and [math]c \neq 0[/math] so the only way this can happen is if [math]p=0[/math], but this is the initial photon momentum, which isn't 0 by definition, which makes this situation nonsensical.

 

 

So, I do understand that a photon can't be completely absorbed by the electron judging by the conservation of energy. What I still disagree with is the statement that seems to be assumed from the question of the homework, that just because the final energy of my hypothesized emitted photon is 0, the entire situation is nonsensical. It just seems like bad physics-thinking to me.

 

If we know in advance that the energies - initial and final - are unequal, and therefore can't be fully absorbed, that's one thing. But if I *only* take into account that the final photon has no energy, on its *own*, this statement can lead to 2 conclusions: either the entire process is nonsensical, *OR* there is no emitted photon.

 

It seems to me that the professor expected us to assume nothing in the beginning (we weren't asked to equate the full initial/final energy conservation calculation) and just judging by the fact that a final emitted photon can't exist, we should conclude that the process can't happen. Isn't that backwards?

 

I am being a bit pedantic here, but I am just trying to explain why I had such a problem with the request to set the photon energy to 0, and why it made no sense. The other exercise with the frames of reference (essentially asking whether an electron can "pop-out" a photon, which has larger energy and larger momentum out of it) makes much more sense in the judgment that this can't occur. The E=0 situation just assumes we have prior knowledge that the question doesn't give yet...

 

Does that make sense?

Edited by mooeypoo
Link to comment
Share on other sites

The equation is a statement of conservation of energy and momentum. If it were possible to absorb the photon, the trig function would be one that diverged at some angle, like tan(theta). Since it doesn't, you can conclude a photon must be emitted.

 

One thing this shows is that there are often multiple ways of approaching a physics problem. You do what works for you.

Link to comment
Share on other sites

Yeah.. I think I'm starting to get the point, though. The equation I was dealing with had both lambda0 (initial photon) and lambda1(emitted photon), so the 'impossibility' was related to the emitted photon too. What I was having troubles with at first is thinking that the equation is only about the original photon and I didn't understand how we can make a determination about the emitted photon too based on that function alone...

 

But yeah, I can see it now, I think. The equation is connecting BOTH the original and the emitted photon, that's why the end result being nonsensical can be related to the emitted photon.

 

This, as much as it was frustrating, was actually quite a learning experience.

 

Thanks, guys (ajb and timo on the chatroom and swansont on thread) for bearing with me on this one :)

 

~moo

Link to comment
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.