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Energy released in particle decay w/ relativity

  1. Feb 25, 2009 #1
    One of the possible decay modes of the neutral kaon is K-> pion + pion The rest energies of the K0 and pion are 498 MeV and 135 MeV, respectively. The kaon is initially at rest when it decays.

    a) How much energy is released in the decay?

    b) What are the momentum and relative direction of the two neutral pions


    Possible equations:
    E0=mc^2
    E=(gamma)mc^2
    p=(gamma)mv



    I got logged out of my last attempt at a post, so let me see if I can replicate it ;x

    I figured at first that the energy released would deal with the loss of energy from the 498 MeV = 135MeV+135MeV +???, but these are all rest energies. I'm not sure if there is a difference, but if the energy released by this was not 228 (498-170) MeV then I would have to delve into the wonderful world of relativity. However, I am not sure of what options I have here. Multiplying it by gamma seems too easy. :/

    For momentum we take p=mv(gamma). We need to find gamma and v. E0=mc^2 gives us the mass (m=E0/c^2). E=(gamma)mc^2 throws us another nice equation using these variables. We can plop E0=mc^2 in there to target the gamma. We then get E=(gamma)E0. or (gamma)=E/E0. But now we need to find E also.

    Any help would be appreciated
     
  2. jcsd
  3. Feb 27, 2009 #2

    malawi_glenn

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    First to energy conservation:

    E_initial = E_final

    Gives you

    [tex]m_Kc^2= \gamma m_\pi c^2 + \gamma m_\pi c^2 [/tex]

    "Energy released" is the mass difference, i.e. the total amount of kinetic energy. That is a bit tricky, since it is not a fundamental quantity. In reality, energy is not released, it is conserved.. E = rest_mass + kinetic_energy

    From the equation above, the E_final = E_initial, you solve for \gamma

    Using that the kaon is at rest, the momenta of pion must be back to back, i.e

    E_pi 1 = E_pi 2

    Use also E^2 = (pc)^2 - m^2c^4

    Now make sure you understand these hints, and show us what you obtain and how you do it.
     
  4. Feb 27, 2009 #3
    It's true that Ei=Ef but there should be energy released in the form of heat, sound etc and I believe that is what I should be looking for. The Epi+Epi would equal the Ekaon? Working with this in mind I started working on finding gamma
    Ei=Ef
    mkc^2=2*(gamma)mpic^2
    Ek=mkc^2
    mk=Ek/c^2=498MeV/c^2
    Epi=mpic^2
    mpi=Epi/c^2=135MeV/c^2

    so: (498MeV/c^2)*c^2=2*(gamma)(135MeV/c^2)*c^2 = 498MeV=370*(gamma)
    (gamma)=1.844
    mk=2mpi*1.844
    So the difference in mass between the two is by a factor of 1.844

    define: KE=((gamma)-1)mc^2
    KE=.844*(2mpi*c^2)
    KE=227.998MeV/c^2

    We have
    E0=498
    KE=228

    I would figure I could use E=E0+KE for this, but that doesn't seem to apply. Am I on the right tracK?
     
  5. Feb 27, 2009 #4

    malawi_glenn

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    energy is "released" in form of kinetic energy of the pions.

    But the energy of the pion is not:

    Epi=mpic^2, totally unnecessary to even write this line and mpi=Epi/c^2=135MeV/c^2 !!

    gamma = 1.8444, yes, that is correct answer

    The kinetic energy can more easily be evaluated from m_k = 2m_pi + KE = E_final ... but yes, 228MeV is correct, you should not have 227.998MeV/c^2, which is wrong units!!!

    Since you had E_K = 498MeV = 2E_pi, you can evaluate what E_pi is, and you have the relation:
    [tex]E^2 = (pc)^2 + (mc^2)^2 [/tex]

    which I gave as hint in the last post, why not try to use it?!
     
  6. Feb 27, 2009 #5
    E=KE+E0
    E=228+498
    E=726

    726^2=p^2c^2+498^2

    279072=p^2c^2
    3.1*10^-12=p^2
    p=1.76*10^-6

    I took E to be the kinetic energy released plus the original value of the kaon at rest. Logically I don't quite understand that though since the KE is energy that was taken from the E0.
     
  7. Feb 27, 2009 #6

    malawi_glenn

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    I dont think you know how to use the equation...

    The energy-momentum-mass relation for a particle, with rest mass m is:

    E^2 = (pc)^2 + (mc^2)^2

    where p is the particle momentum and E is the energy. I thogut that was clear, but it wasn't ...

    Now you wanted to find the momentum of the pion, go ahead and try again. Use that the energy of the pion must the (1/2) of the Kaons rest-mass energy.

    Of course you cannot logically understand what you did since what you did is rubbish/nonsense.
     
  8. Feb 27, 2009 #7
    Yes the downside of missing sleep ; ; I was mixed up on the rest energy there.
    E0=135=mc^2
    KE=114
    E=249

    Those are the real values now lol

    E^2=p^2c^2+(mc^2)^2
    249^2=p^2c^2+135^2
    249^2-135^2=(pc)^2
    43776=(pc)^2
    p^2=43776/c^2

    sqrt everything

    p=209MeV/c
     
  9. Feb 27, 2009 #8
    Thank you for all your patience and help it's been much appreciated :D
     
  10. Feb 27, 2009 #9

    malawi_glenn

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    Ok, great well done
     
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