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Homework Help: Electron/positron and photon

  1. Apr 3, 2014 #1
    1. The problem statement, all variables and given/known data

    A gammaphoton with a frequency of 5,49*10^20 Hz passes by near a nucleus, and turns into a electron/positron pair. The electron has an kinetic energy of 1,20*10^-13 J. How much kinetic energi is in the posistron?(we can neglect the energyexchange with the nucleus)

    3. The attempt at a solution

    Ive tried to solve this using this formula but i get a negative answer:

    hf=2E+Ekp+Eke => Ekp=hf-2E-Eke

    Where Ekp is the kinetic energi of the positron, "hf" is the energy of the photon, "E" is the rest energy of the electron and "Eke" is the kinetic energy of the electron.

    I get that Ekp=-1,64*10^-10 J. This isnt right?
  2. jcsd
  3. Apr 3, 2014 #2
    The formula is correct. You must've made a mistake in your calculation.
  4. Apr 3, 2014 #3


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    Homework Helper

    your mistake is in the Units ... 1 Joule is 1 kg m2/s2 , so your electron mass must be in kg.
  5. Apr 4, 2014 #4
    As far as i can understand, the energy in the photon is not enough to create a electron/positron pair?


    And therefore the problem doesnt have an answer?
  6. Apr 4, 2014 #5


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    You're wrong, the problem does have an answer, which I just found. So we can be sure the problem is correct.

    Your equation is also correct, so the error must lie in your working.

    Unless you show this working, in detail, noone can help you further.
  7. Apr 4, 2014 #6

    Ekp=6,63*(10^-34)Js*5,49*(10^20)Hz - 2*9,1094*(10^-31)kg*(3,00*(10^8)m/s)^2 - 1,20*(10^-13)J


    I just kept entering the wrong exponent in C^2.

  8. Apr 4, 2014 #7


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    What I got too.

    Here's a tip: Google calculator is your friend. Take a look at this: https://www.google.com.sg/search?q=...+-+2*mass+of+electron*c^2+-+1.2E-13J&safe=off

    Just enter your expression as a query next time and Google works everything out, physical constants, units, etc. all inclusive!
  9. Apr 5, 2014 #8
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