# Relativistic energy and momenta

1. Jun 20, 2009

### fredrick08

1. The problem statement, all variables and given/known data
An electron with a kinetic energy of 1MeV collides with a stationary positron. The two particles annihilate each other and produce 2 photons, of equal energy traveling at angle theta to the direction of the electron. Find,

a. The momenta of the electron
b.the energy of the photons
c.the momentum of the photons
d. the angle theta of the photons

rest mass electron/positron = 0.511MeV/c^2
speed of light c = 3x10^8 m/s
E^2=(Eo+Ek)^2=p^2*c^2+(mc^2)^2

3. The attempt at a solution
ok for a. which is all i have done so far im pretty sure all i have to do is solve for p

p=root((((Eo+Ek)^2)-((mc^2)^2))/c)=??? i am getting very confused with the units in this question ..

I am getting 1.422MeV/c please can any tell me if this is right?

2. Jun 20, 2009

### fredrick08

well b. would be Ef=Ei=Ek+Eo=2Ephoton=>Ephoton=(Eo+Ek)/2=0.75MeV

3. Jun 20, 2009

### fredrick08

ugh c. momentum of photons... they dont have mass so no rest energy? therefore there momentum would the same as their energy divided by c? 0.75MeV/c

4. Jun 20, 2009

### fredrick08

d. no idea how to find the angle....

5. Jun 20, 2009

### fredrick08

any help for any question would be very much appreciated thankyou.

6. Jun 20, 2009

### fredrick08

arccos(0.53/0.75)=45 degrees?? makes sense

7. Jun 20, 2009

### superkan619

The diagram according to me is like:

http://www.geocities.com/vinteract11/derytuio.jpg

The outgoing photons make equal angle thita with initial direction of e- and hence the final momentum is along the initial momentum. Furthermore the momentum (hence energy) of both the photons need to be same in magnitude. For each momentum you have to use vector resolution along X and Y axes. Am I correct?

Last edited: Jun 20, 2009
8. Jun 20, 2009

### fredrick08

yes i think so, that is what i did... i think lol

9. Jun 20, 2009

### superkan619

Correct eqn is:
p=root( [((Eo+Ek)^2)/c^2]-[(mc^2)^2))/c^2] )

Which gives p = (root2) MeV/c

10. Jun 20, 2009

### superkan619

Energy of each photon = (2Eo + Ek)/2 ;since the rest energy of positron(Eo) too is converted into photonic energy.
This gives each photon has energy 1.011 MeV.
So Momentum of each photon is 1.011 MeV/c.

11. Jun 20, 2009

### fredrick08

ahh ok thankyou

12. Jun 20, 2009

### fredrick08

ok then for d... what do i have to do to find the angle?

13. Jun 20, 2009

### fredrick08

have i done it right? if i use 1.011 i get 44.3 degree?

14. Jun 20, 2009

### fredrick08

sorry lol... miss typed.. ok i see now is always root2, so 45 degrees is correct, thnank very much

15. Jun 20, 2009

### superkan619

Angle stinks!!