Relativistic energy and momenta

[fredrick08]

Homework Statement

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

The Attempt at a Solution

ok for a. which is all i have done so far I am 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?

 

[fredrick08]

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

 

[fredrick08]

ugh c. momentum of photons... they don't have mass so no rest energy? therefore there momentum would the same as their energy divided by c? 0.75MeV/c

 

[fredrick08]

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

 

[fredrick08]

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

 

[fredrick08]

how about for d. E^2=2B^2=>B=(root(2)*E)/2=0.53

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

 

[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?

 

[fredrick08]

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

 

[superkan619]

The Attempt at a Solution[/h2]
ok for a. which is all i have done so far I am 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?

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

Which gives p = (root2) MeV/c

 

[superkan619]

well b. would be (Eo+Ek)/2=0.75MeV

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.

 

[fredrick08]

ahh ok thankyou

 

[fredrick08]

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

 

[fredrick08]

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

 

[fredrick08]

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

 

[superkan619]

Angle stinks!