# Pair production problem

1. Feb 7, 2007

### kreil

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

A photon of energy E strikes an electron at rest and undergoes pair production, producing a positron and an electron:

photon + (e-) = (e+) + (e-) + (e-)

The two electrons and the positron move off with identical momenta in the direction of the initial photon. Find the kinetic energy of the three final particles and find the energy E of the photon.

2. Relevant equations

1.$$K=E-E_0$$

2.$$E_0=mc^2$$*rest energy of photon = 0.

3.$$E=\frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}$$

4.$$E=\sqrt{(pc)^2+(mc^2)^2}$$

5.$$p_i=p_f=\frac{mv}{\sqrt{1-\frac{v^2}{c^2}}}$$

3. The attempt at a solution

The initial momentum of the photon is mc, and this gives an initial kinetic energy of 2mc2. This is equal to the final kinetic energy, and since each particle has the same momentum then the kinetic energy of each is 2mc2/3.

My problem comes with finding the energy of the photon. The book gives an answer of 4mc2, and I'm not sure how to get that. Any help is appreciated.

Josh

2. Feb 7, 2007

### Tom Mattson

Staff Emeritus
Let's start right here:

No, it isn't. The mass of a photon is zero, so that can't be its momentum. Use your equation 4 to determine the photon momentum in terms of the photon energy E.

3. Feb 7, 2007

### kreil

$$E=\sqrt{(pc)^2+(mc^2)^2} \implies p=\sqrt{\frac{E_p^2}{c^2}}$$

..what do i do with this?

Last edited: Feb 7, 2007
4. Feb 7, 2007

### Tom Mattson

Staff Emeritus
Take the square root. You get $p=E/c$. That's the momentum of the photon in terms of its energy.

Now let's go on to the next part.

I have no idea of how you got 2mc2, but if you didn't use the conservation laws then you didn't do it correctly. Also, it is not the case that all of the photon energy goes into the KE of the 3 leptons. Some of the photon energy had to be used to create the pair!

So, can you write down the laws of conservation of energy and momentum in terms of the symbols that were given to you, and the final lepton speed v? Note that I can say that they are all moving at the same speed, because they all have the same momentum and mass.

5. Feb 7, 2007

### kreil

$$p_i=p_f=3p_e=\frac{3mv}{\sqrt{1-\frac{v^2}{c^2}}}=3mc$$

6. Feb 7, 2007

### Tom Mattson

Staff Emeritus
That's the momentum of each of the final particles. That is not a conservation law. The law of conservation of X (X=energy or momentum) is that the total amount of X in a closed physical system is the same before and after an interaction. You know that 2 quantities are conserved here, so you should be able to write down 2 equations.

The momentum of the photon before the pair production is_____?
The momentum of the electron before the pair production is_____?
The momenta of each of the 3 particles after the pair production is____?

Earlier in this thread, I told you the answer to the first one. You just told me the answer to the third one.

So, what's the answer to the second one? And can you construct the conservation equation from these pieces of information?

Try that, and then we'll move on to conservation of energy.

7. Feb 7, 2007

### kreil

The momentum of the electron is mc, so

$$\Sigma p_i=\Sigma p_f \implies \frac{E_p}{c}+mc=3mc$$

8. Feb 7, 2007

### Tom Mattson

Staff Emeritus
No. Here's a hint: The electron is at rest before the pair production. (How did you get mc by the way?). And the final momentum isn't 3mc. I have no idea of how you got that either, especially when you wrote down the correct answer in your previous post!

9. Feb 7, 2007

### kreil

oops! So the initial electron has no momentum. And I thought I did write down 3mc in my previous post, as thats where I got it from...

10. Feb 8, 2007

### Tom Mattson

Staff Emeritus
Jeez, I didn't even notice that you wrote 3mc in that post. No, the correct answer for the particle momenta after the pair production is what comes before the 3mc:

$$3p_e=\frac{3mv}{\sqrt{1-\frac{v^2}{c^2}}}$$

That does not equal 3mc.