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Find energy threshold from a proton collision

  1. Oct 11, 2013 #1
    1. The problem statement, all variables and given/known data
    Two photons collide and create an electron-positron pair:

    y1+y2 -> e-+e+

    If the wavelength of y1 is 1 mm calculate the energy threshold for photon y1 to produce the electron-positron pair.

    Suppose E(y1)=hc/[itex]\lambda[/itex]2 where [itex]\lambda[/itex]2 =1.1 mm

    Hints: In the center of mass frame of reference, in which the total momentum is 0, the threshold reaction would just produce the two particles at rest. Then in the frame with photon energies E1
    and E2 , the two particles would have the same momentum.

    2. Relevant equations
    Given:
    Electron Energy=(p2c2+m2c4)1/2
    Photon Energy=pc where p=momentum


    3. The attempt at a solution
    The energy of y2 is planck's constant*c/1.1 mm = 1.81*10-22 Joules

    I'm a bit confused about where to go from here. Seeing as how energy and momentum are conserved the total energy of y1 and y2 will have to equal the energy of the electron-positron pair. Since I have E(y2) and need to find E(y1) I need to find the energy of the electron-positron pair.

    I'm given the energy equation for the electron and their mass, c, and p should all be equal so their energies will be equal.

    Is the following assumption correct?
    E(y2)=2*(p2c2+m2c4)1/2-E(y1)

    The problem is I don't have a wavelength to figure the energy of the resulting electron-positron pair.

    Any suggestions?

    Thank you!
     
  2. jcsd
  3. Oct 11, 2013 #2

    mfb

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    Your indices (1, 2) look mixed up.
    It is.

    Momentum conservation will give you the second equation, so you can solve for both the missing photon energy and the electron/positron momentum at the same time.
     
  4. Oct 11, 2013 #3
    Ah yes, thank you!:

    E(y1)=2*(p2c2+m2c4)1/2-E(y2)

    Is this something along the lines of what you were suggesting?:
    my1vy1+my2vy2=me1vy1+me2vy2

    That might not make sense seeing as how the total momentum in the center of mass frame on either side of the reaction should be 0 but outside that frame this should be useful. I'm being thrown a bit by the transition between frames.
     
  5. Oct 12, 2013 #4

    mfb

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    Photons do not have mass, and you will need the relativistic momentum for the electrons.
    Consider momentum conservation in our lab frame, not in the center of mass frame.
     
  6. Oct 12, 2013 #5
    Ah, not sure what I was thinking there. Thank you!

    E(y1)/c+E(y2)/c=me1vy1+me2vy2

    Do I instead have to solve the given electron energy equation for momentum to find the relativistic momentum of the electrons or will the newtonian estimate suffice?

    The mass and velocity of the two electrons should be the same:

    E(y1)/c+E(y2)/c=2meve

    I'm not sure it matters how I represent the momentum of the electrons, actually. It will ultimately be 2*(some value) or 2p. That will be the same value for p that I use in the conservation of energy equation:


    E(y1)=2*(p2c2+m2c4)1/2-E(y2)

    If I solve for p in my momentum equation:

    E(y1)/2c+E(y2)/2c=p

    Plug p in to my momentum equation:
    E(y1)=2*((E(y1)/2c+E(y2)/2c)2c2+m2c4)1/2-E(y2)

    Plug in the known values and solve for E(y1)?
     
  7. Oct 12, 2013 #6
    Doing that I get E(y1)=1.346*10-22 J

    That is at least a reasonable answer. I'm just not sure I can do it that way...
     
  8. Oct 13, 2013 #7

    mfb

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    The second photon needs much more energy than that. Your added photon energies are not even close to the rest energy of an electron, even without the necessary kinetic energy this cannot work out.

    You cannot use the nonrelativistic electron momentum, it gives wrong results. And one of your photon momenta points in the wrong direction.
     
  9. Oct 13, 2013 #8
    Ah, thank you. I thought I swapped out for a relativistic version of momentum but apparently I was wrong! :)

    I guess then that solving the given electron energy equation above (given) for momentum and going from there is what I need to do?
     
  10. Oct 13, 2013 #9

    mfb

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    That is one of many options to solve the system of equations.
     
  11. Oct 13, 2013 #10
    Where is the momenta in the wrong direction?

    Conservation of Energy:
    E(y1)=2(p2c2+m02c4)1/2-E(y2)

    Conservation of Momentum:
    I know that the energy of a photon = p*c so p=[itex]\frac{E}{c}[/itex]

    The sum of the photon momentum should equal the sum of the electron momentum.

    I know that the energy of an electron = mc2=(p2c2+m02c4)1/2

    Solving this for p I get:

    p=[itex]\sqrt{c^2(m^2-m_0^2)}[/itex]

    But now I've introduced another variable, m0, which is the resting mass of the electron and m is the relativistic mass (unknown):

    E(e)=mc2 so:

    m=E(e)/c2 but I'm again referencing the unknown energy of the electron. I'm not sure where to go from here...

    Anyway, here is what the momentum for one of my electrons would be:
    sqrt(c2([itex]\frac{E^2}{c^4}[/itex]-m02))

    How do I cancel out this additional unknown?

    Can I use E=mc2 but use the total energy of the two photons to solve for the total relativistic mass of the two electrons and then divide by two to get the individual mass?

    If that is the case:

    E(y1)+E(y2)/c2=metotal

    Suggestions? Thanks again
     
  12. Oct 13, 2013 #11

    mfb

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    That is the absolute value of the momentum. If your photons travel in opposite directions, the momenta point in opposite directions, and you have to subtract their momenta.

    Relativistic mass is a bad concept, try to avoid it.
     
  13. Oct 13, 2013 #12
    The only way I see so far to avoid it is to skip the mc2 portion and just say E(e)=(p2c2+m02c4)1/2 and solve for p but in that case I don't know how to get rid of the E(e) unknown because we're not talking about a resting electron. Is there a third equation I should be using?
     
  14. Oct 13, 2013 #13

    mfb

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    No third equation.

    This collection of formulas gets messy. Can you just write down energy and momentum conservation in terms of the photon energies Ey1, Eyy2 (and please make clear which of them you know), p of electron (=p of positron) and constants like the electron mass and the speed of light?
     
  15. Oct 13, 2013 #14
    Absolutely!

    Conservation of energy:
    E(y1)=2(p2c2+m02c4)1/2-E(y2)

    I know Ey2=planck's constant*c/1.1 mm = 1.81*10-22 Joules
    c=speed of light (known)
    m0=mass of resting electron (known)

    The only unknowns in this equation are the momentum, p, of the electrons and Ey1.

    Conservation of momentum:
    I believe this is where I have to switch the direction of my photon momentum which is a bit confusing because the electrons would also be traveling in opposite directions (?) but should be equal and would thereby add up to 0 but because they're not zero and we have two then it will be 2*p. Anyway, we can get p from the energy of an electron equation:

    E(e)=(p2c2+m02c4)(1/2)

    Solving for p=([itex]\frac{E_(e)^2-m_0^2c^4}{c^2}[/itex])(1/2)

    Again, I know m0 and c. I don't know E(e). I'm guessing there is a different way to write this but I don't know it.

    The equation for conservation of momentum:
    ([itex]\frac{1}{c}[/itex])(Ey1-Ey2)=2*(([itex]\frac{E_(e)^2-m_0^2c^4}{c^2}[/itex])(1/2))
     
    Last edited: Oct 13, 2013
  16. Oct 13, 2013 #15

    mfb

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    The energy conservation is good.

    At threshold, electron and positron will be at rest relative to each other (so in the center of mass frame, they have no kinetic energy). That means that in the lab frame, they move in the same direction.

    Please do not try to do multiple steps at the same time if you have problems with individual steps. It is easier if you follow my advice and use p here.

    The equation for momentum conservation looks good.

    If you want to use energy: fine. You can use the conservation of energy to express E(e) in terms of the photon energies, then the equation has just Ey1 as unknown quantity.
     
  17. Oct 13, 2013 #16
    Oh, no, I have no preference at all. I just thought that is what you were suggesting I do. Sorry for the confusion!

    Where? Do you mean just leave the momentum conservation equation as:

    ([itex]\frac{1}{c}[/itex])(Ey1-Ey2)=2*p

    The only other way I know how to represent momentum of the electrons is to solve that energy equation. I don't know of a better way unless I can just leave it as p flat out.

    EDIT: Other than the fact that I changed the direction of my momentum this is effectively what I tried earlier where I came up with an answer that we agree was too small for Ey1. If we simply leave it as p it doesn't matter what p is because we just plug it into the energy equation and p goes away.
     
  18. Oct 13, 2013 #17

    mfb

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    That's the idea.

    If you get a wrong value, it would be useful to show how you simplified the equation and finally calculated the photon energy.
     
  19. Oct 13, 2013 #18
    Great, well, if we leave it as p in the momentum equation we can solve for p easily:

    p=[itex]\frac{(1/c)(E_1-E_2)}{2}[/itex]

    Now we have our energy equation:

    E1=2*[itex]\sqrt{p^2c^2+m_0^2c^4}[/itex]-E2

    We can take the p we found from the momentum equation and plug it into this energy equation:
    Note- The notation is easier if I just use E1 and E2 vs Ey1 and Ey2

    E1=2*[itex]\sqrt{[(1/(2c))(E_1-E_2)]^2c^2+m_0^2c^4}[/itex]-E2

    (Inside the square brackets is p)

    Take the 1/2c out of the brackets after squaring it and cancel the c^2's:
    E1+E2=2*[itex]\sqrt{((1/4)(E_1-E_2))^2+m_0^2c^4}[/itex]

    Square both sides and multiply through by the new 22:
    (E1+E2)2=(E1-E2)2+4m02c4

    Expand the two squares:
    E12+E22+2E1E2=E12+E22-2E1E2+4m02c4

    Consolidate/cancel the E's:
    4E1E2=4m02c4

    E1=[itex]\frac{m_0c^4}{E_2}[/itex]

    Which works out to E1=3.71351*10-5 J

    Seems like a lot!
     
    Last edited: Oct 13, 2013
  20. Oct 14, 2013 #19

    mfb

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    That looks good.
     
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