Electron-positron collision producing photons at 0.60c

[Phymath]

what the !?

ok idk if I am right and the book is wrong but help me out the question is...

An electron moving at a speed 0.60c collides head on witha positron also moving at 0.60c. Determine the energy and momentum of each photon produced in the process.

well this is what I did..
$$E = 2\frac{m_ec^2}{ \sqrt{1-\frac{(0.6c)^2}{c^2}}}$$
$$E = 255864.36 \ eV = 0.256 \ MeV$$
the book says the energy is E = 0.64 MeV why?

i figured E = m_e c^2 + KE but it my calcs must be off or something i don't understand

 

[Integral]

you are not using the correct velocity. Since both electrons are moving at .6c you must compute the speed of the collision using the relativistic formula for addition of velocities.

 

[Fredrik]

Integral: That was my first thought too, but no one said we're supposed to do the calculation in the electron's rest frame. He is doing the calculation in the center-of-mass frame, which is OK.

Phymath: That square root is 0.8, so the quantity you call E (the sum of the two energies), is just 2.5 times the rest mass of the electron, which is 0.511 MeV. So it looks like you just did a very simple calculation error. 2.5 times 0.511 is not 0.25.

 

[Hurkyl]

Right, google gives 1 277 497.26 electron volts for that calculation

http://www.google.com/search?hl=en&...+*+(speed+of+light)^2+/+0.8+in+ev&btnG=Search

 

[Fredrik]

Wow, that was cool! I had no idea you could do that.

 

[Phymath]

i understand the .8 error but still it does not give the correct energy value...

 

[Fredrik]

Divide it by 2. There are 2 photons. (They must have the same energy because the total momentum is 0 in this frame).