How Is the Total Energy of Photons Calculated in Electron-Positron Collisions?

[magma_saber]

Homework Statement

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?

Homework Equations

mass of an electron = 9e-31 kg
E=$$\gamma$$mc² - mc²

The Attempt at a Solution

2.2032e-13 - 8.1e-14 = 1.3932e-13
1.393e-13 * 2 = 2.7864e-13 J

 

[Redbelly98]

I have a few comments:

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.

 

[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

 

[Redbelly98]

the questions ask for in Joules.

Okay, understood.

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

Looks good.

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

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.

Then i multiplied it by 2 since that is the energy of one of the particle.

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

 

[Yokushin]

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.

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?