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Calculate muon speed after collision

  1. May 6, 2015 #1
    1. The problem statement, all variables and given/known data

    Two equal energy photons collide head on and annihilate each other, producing a u+, u- pair. The two particles have equal mass, about 207 times the electron mass.

    A) Calculate the maximum wavelength of the photons for this to occur.
    B) If the wavelength calculated in A) is halved, what is the speed of each muon after they have moved apart (use the correct expressions for relativistic momentum and energy)

    2. Relevant equations

    E = hc/lambda

    3. The attempt at a solution

    I have solved part A) using the equation above, with the maximum wavelength occurring for the minimum energy (the rest energy of the muon pair)

    lambda max = hc/((3.394x10^-11)/2) = 1.17x10^14m

    I am having trouble finding the speed of each muon after they have moved apart...

    lambda = 5.855x10^-15 for each photon (from A))
    then each photon has E = 3.394x10^-11J

    So through conservation of energy using E^2 = (pc)^2 + (m0c^2)^2

    6.788x10^-11 = 2(pc + m0c^2)

    p = 5.656689x10^20

    using p = (mv)/(1-v^2/c^2)^(1/2) rearrange to find p^2=v^2(m^2+(p^2/c^2))

    rearrange to find v^2 and input the values of p, m and c where m = 207(9.109x10^-31)

    gives v = 0.709c where c is the speed of light

    This answer is wrong, but I can not see where I have made a mistake in calculation.
     
  2. jcsd
  3. May 6, 2015 #2

    PeroK

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    Two things first:

    You're doing "advanced physics" but you have a "high school" approach of putting complicated numbers into equations at the first opportunity. You ought to work at changing that. You should be doing a problem like this algebraically, then plugging in the numbers at the end.

    You ought to learn some latex. It's difficult to read what you've written.

    In terms of a solution, one of the advantages of an algebraic approach is that you can see the physics instead of a lot of numbers to 6 decimal places.

    This is a good oppotunity to leave the "high school" approach behind. I'll get you started:

    Let ##\lambda## be the maximum wavelength and consider two photons with wavelength ##\lambda/2##.

    The total energy of the two photons is:

    ##E = 4hc/\lambda##
     
  4. May 6, 2015 #3

    Orodruin

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    Try to first find the gamma factor of the muons. How much is the energy of each muon? How does a particle's energy and mass relate to the gamma factor?
     
  5. May 8, 2015 #4
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    Last edited: May 8, 2015
  6. May 8, 2015 #5
    ....................................
     
    Last edited: May 8, 2015
  7. May 8, 2015 #6

    Orodruin

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    If that is the relation to the kinetic energy, what is the relation to the total energy?

    Yes, this is correct, but the argument can be made much simpler if you consider the total energy for one muon.
     
  8. May 8, 2015 #7
    Indeed it does - thanks! So the velocity calculated is the speed of each muon after the collision?
     
  9. May 8, 2015 #8

    Orodruin

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