# Sum of photon energies

1. Feb 16, 2009

### magma_saber

1. The problem statement, all variables and given/known data
Suppose that a positron traveling at a speed of 0.93c collides head on with an electron traveling at the same speed.

What is the sum of the energies of the two photons?

2. Relevant equations
mass of an electron = 9e-31 kg
E=$$\gamma$$mc2 - mc2

3. The attempt at a solution
2.2032e-13 - 8.1e-14 = 1.3932e-13
1.393e-13 * 2 = 2.7864e-13 J

2. Feb 16, 2009

### Redbelly98

Staff Emeritus

It's kind of hard to follow what you are doing. Including units with all values in your equations would help with that.

I don't see where or if you have calculated γ.

Also, it's probably easier to work in eV energy units, rather than Joules. Your textbook might even give the value of mc2 for the electron, in eV units.

3. Feb 17, 2009

### magma_saber

the questions ask for in Joules.

rest energy = mc2
(9e-31)*(3d8)2 = 8.1e-14

$$\gamma$$ = 1/$$\sqrt{1-(0.93)^2}$$ = 2.72

particle energy = $$\gamma$$*mc2
(2.72)*(9e-31)*(3d8)2 = 2.20e-13

E = particle energy - rest energy
E = 2.20e-13 - 8.1e-14 = 1.39e-13

Then i multiplied it by 2 since that is the energy of one of the particle.
1.39e-13*2 = 2.79e-13 J

4. Feb 17, 2009

### Redbelly98

Staff Emeritus
Okay, understood.

Looks good.
Here E is the kinetic energy of one of the particles. However, it is the total particle energy, kinetic and rest mass, that ultimately is converted into the photons.

Yes, but you'll need to use the total energy rather than only kinetic. Looks good otherwise.

5. Nov 14, 2009

### Yokushin

I have a similar problem and I followed the posters procedure but I am however confused as to why we need to add rest mass as the instructor has said. If someone would clarify this problem in better detail much appreciated.

I am just confused, is the final formula:
2(RE + KE + PE + Mass) = Total Energy of the two photons?