1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Relativistic energy and momenta

  1. Jun 20, 2009 #1
    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. jcsd
  3. Jun 20, 2009 #2
    well b. would be Ef=Ei=Ek+Eo=2Ephoton=>Ephoton=(Eo+Ek)/2=0.75MeV
     
  4. Jun 20, 2009 #3
    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
     
  5. Jun 20, 2009 #4
    d. no idea how to find the angle....
     
  6. Jun 20, 2009 #5
    any help for any question would be very much appreciated thankyou.
     
  7. Jun 20, 2009 #6
    how about for d. E^2=2B^2=>B=(root(2)*E)/2=0.53

    arccos(0.53/0.75)=45 degrees?? makes sense
     
  8. Jun 20, 2009 #7
    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
  9. Jun 20, 2009 #8
    yes i think so, that is what i did... i think lol
     
  10. Jun 20, 2009 #9
    Correct eqn is:
    p=root( [((Eo+Ek)^2)/c^2]-[(mc^2)^2))/c^2] )

    Which gives p = (root2) MeV/c
     
  11. Jun 20, 2009 #10
    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.
     
  12. Jun 20, 2009 #11
    ahh ok thankyou
     
  13. Jun 20, 2009 #12
    ok then for d... what do i have to do to find the angle?
     
  14. Jun 20, 2009 #13
    have i done it right? if i use 1.011 i get 44.3 degree?
     
  15. Jun 20, 2009 #14
    sorry lol... miss typed.. ok i see now is always root2, so 45 degrees is correct, thnank very much
     
  16. Jun 20, 2009 #15
    Angle stinks!!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Relativistic energy and momenta
  1. Relativistic energy (Replies: 2)

  2. Relativistic energy (Replies: 1)

  3. Relativistic energy (Replies: 1)

  4. Relativistic energy (Replies: 1)

Loading...