# Pion decaying into 2 photons

1. Apr 30, 2009

### cashmoney805

1. The problem statement, all variables and given/known data
The neutral pion is an unstable particle that decays very quickly after its creation into two photons (“particles” of light: v = c, mo = 0). The pion has a rest-mass energy of 135 MeV. Consider a pion that has a kinetic energy of 90 MeV
1) Determine the v of this pion
2) Determine the momentum of the pion
3) Determine the sum of the energies of the photons

2. Relevant equations
1) moc2$$\gamma$$ = KE + Rest E
2) mov$$\gamma$$ = p
3) Ephotons= pc

3. The attempt at a solution
I'm pretty sure I got 1 and 2. For 1 I added the KE + RE, converted to proper units, and finally got v = .80c
For 2 I got p = 9.6 x 10-20 kg m/s
Now I'm not so sure about #3. I think Ephotons = pc because momentum is conserved, and all of the pion's momentum gets turned into the photons' momentum. However, what about adding up the original rest mass energy + KE of the pion? If I do that I get a bigger energy then if I do pc. Thanks for all the help!

2. Apr 30, 2009

### Nabeshin

Just using pure energy conservation, how much energy is there before and after the decay?

3. Apr 30, 2009

### cashmoney805

According to my calculations, E before = (90 + 135) MeV = 225 MeV
After E = pc = 181.25 MeV

4. May 1, 2009

### cashmoney805

I calculated p a different way this time, p = sqrt(2mKE) where m is the relativistic mass. When I do this then multiply p by c to get E, I get E = 201 MeV. Oh boy...

edit: actually I'm not sure if that equation works for speeds close to c...

Last edited: May 1, 2009
5. May 1, 2009

### Nabeshin

Hrm, you're letting the calculations get bogged down too much. The change in energy is zero, right? You know initial energy, so final energy must be.. the same.

6. May 1, 2009

### cashmoney805

I get what you're saying, but I don't understand why the equations don't work here. It seems to me that momentum isn't conserved.

Last edited: May 1, 2009
7. May 1, 2009

### cashmoney805

and there is one more part to this problem which I thought I could get myself, but I can't. Here is a pic of the question/diagram

http://i44.tinypic.com/k12kja.gif