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Pair Production and kinetic energy

  1. Jul 16, 2003 #1

    A photon of energy E strikes an electron at rest and undergoes pair production, producing a positron and another electron.

    photon + e- --------> e+ + e- + e-

    The two electrons and the positron move off with identical linear momentum in the direction of the initial photon. All the electrons are relativistic.

    a) Find the kinetic energy of the final three particles.

    b) Find the initial energy of the photon.

    Express your answers in terms of the rest mass of the electron mo and the speed of light c.

    I set the problem up this way:

    For part a)

    E is conserved before and after the collision.

    Ephoton = 3Ee + 3KEe

    hc/&lambda = 3moc2 + 3KEe

    I then solved for KEe and found:

    KEe = 1/3hc/&lambda - moc2

    Is this the correct method to find the kinetic energy of one of the electrons after the collision?

    The problem says "state your answer in the rest mass of the electron and the speed of light", however if I don't know the wave length of the photon, how do I eliminate that energy term. Since they say the electrons are moving relativistic, do I just assume their speed is equal to 1/10c and then work backward to find the photons wavelength?

    I may be way off, any help is appreciated!


    Edit: I need to use mc2(&gamma - 1) for KE!

    The biggest question I have is linear momentum conserved. I would assume it is, and the problem hints at linear momentum.
    Last edited: Jul 16, 2003
  2. jcsd
  3. Jul 16, 2003 #2

    Tom Mattson

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    Not quite. You forgot the rest mass energy of the original electron. Also, what is "Ee" on the right hand side? I presume that's the rest mass energy of each particle.
  4. Jul 16, 2003 #3
    Is this okay?

    Ephoton + moc2= 3moc2 + 3Ke

    Ephoton = energy of the photon: hc/&lambda
    moc2 = rest mass energy of electron
    Ke = kinetic energy of the electrons in motion (all three with equal linear momentum and kinetic energy) after the photon collides with the initially motionless electron.
  5. Jul 16, 2003 #4

    Tom Mattson

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    That's a good start, but you have to take it further. The problem said to express the answer in terms of m0 and c. That means you can't have λ in the answer. So now you need to conserve momentum and get a second equation.

    Also, rather than write the total energy of each final particle as m0c2+K, I would write each one as γm0c2. That will make it easier to simultaneously solve the 2 equations.
  6. Jul 16, 2003 #5
    Is the algebra soposed to be really messy on this problem.[?]

    I just want to make sure I'm not going through all this for nothing.
  7. Jul 16, 2003 #6

    Tom Mattson

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    There is a messy way, and there is a clean way.

    First, the 2 equations.

    hc/λ+mc2=3γmc2 (Conservation of Energy)
    h/λ=3γmv (Conservation of Momentum)

    The clean way is to multiply the second equation by 'c', and move the mc2 to the right side in the first equation. Then, the left hand sides of the two equations are equal, so you can set the right hand sides equal to each other. Solve for v, then get the KE from that. Then use either equation (I'd use the second one because it's shorter) to find the initial energy of the photon.
  8. Jul 16, 2003 #7
    I like your way much better.

    I ended up with a &gamma2/[(&gamma +1)(&gamma-1)] = 4/9mc2, solving for v would have been fun. I bet I made some mistakes along the way, two pages of algebra and I write small. Plus that I tried to solve for K directly, then subsituted that and tried to go for v. It's ugly.

    BTW: I found v to be 2/3c
    Last edited: Jul 16, 2003
  9. Jul 16, 2003 #8

    Tom Mattson

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    It seems you did make a mistake.

    Here are your two equations, after the manipulations I described:

    hc/λ=(3γ-1)mc2 (Conservation of Energy)
    hc/λ=3γmcv (Conservation of Momentum)

    You can straighforwardly solve for v by equating the right hand sides of each equation. For convenience, I would define β=v/c, and I get the following:


    Also note that γ=1/(1-β2)1/2. You should be able to solve for β without much algebra.

    BTW, I got β=0.8, so v=0.8c.

    Edit: fixed math symbol
  10. Jul 16, 2003 #9
    Yeah I found my error. Reworking it right now.

    I canceled &gamma, which you can't do.

    Since E of initial electron is only mc2
  11. Jul 16, 2003 #10
    Are you sure v isn't 8/9c, not .8c.
    Last edited: Jul 16, 2003
  12. Jul 16, 2003 #11

    Tom Mattson

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    I got (4/5)c=0.8c.
  13. Jul 16, 2003 #12
    I get the K of electron after collision:


    and for initial photon energy of:


    If I didn't get those right, I'm quiting for the day.
  14. Jul 16, 2003 #13

    Tom Mattson

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    Those are correct.
  15. Jul 16, 2003 #14
    The easiest way to solve this one

    is backwards. Start in the rest frame of the two electrons and positron, an electron and positron anihilate and the remaining electron absorbs the recoil momentum. Then do a Lorentz transform of the initial and final states.
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