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Momentum of a photon

  1. Feb 7, 2013 #1
    1. The problem statement, all variables and given/known data
    Find the momentum of a 140eV photon.



    2. Relevant equations

    momentum equations

    e=mc^2



    3. The attempt at a solution

    140eV=2.24*10^-17 J

    p=mv
    p=(E/c^2)c
    p=E/c
    p=(2.24*10^-17)/(3*10^8)
    p=7.466666*10^-26




    Is this correct? (other than significant digits)
     
  2. jcsd
  3. Feb 8, 2013 #2

    Dick

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    Your derivation is little suspect since m=0. You should probably really be using E^2=(mc^2)^2+p^2*c^2. But the only problem aside from signficant digits is that you didn't put any units on it.
     
  4. Feb 8, 2013 #3

    Simon Bridge

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    As it happens - yes. But don't forget to write in the units!
    The actual energy-momentum relation is ##E^2=m^2c^4+p_2c^2## and photons have ##m=0##.
    As Dick says, you got the right answer through the (apparently) wrong method.

    Technically you can say that p=140eV/c since "eV/c" are units of momentum :)
     
  5. Feb 8, 2013 #4
    How about this:

    140eV=2.24*10^-34 J

    m=E/c^2
    m=2.488888889*10^-34
    if a photon has no mass (even though the course asks me to find the mass of a photon several times) then how did i get this answer?



    p=h/λ
    mv=h/λ
    mc=h/λ
    λ=.......

    never mind. this is getting roundabout.




    the course has me calculate the mass value for a photon several times.


    when it would ask me for the momentum of a photon, however, it would have me use the formula : p=h/λ
    however, it would also supply a frequency so that i could actually calculate wavelength.

    i don't know of any way to calculate wavelength in this case, so i just used the standard p=mv for a particle
     
    Last edited: Feb 8, 2013
  6. Feb 8, 2013 #5

    Simon Bridge

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    Nope - m=0 for a photon.
    And you left off the units again.

    But you are close (see post #3).
    Mass is a form of energy so there is a sense in which you can attribute a total mass to a photon. It's just potentially misleading to think of it that way.
    i.e. the ##m## in the relation is the rest mass. For a moving mass, the mass-energy relation is ##E=\gamma mc^2##.

    But you do have the frequency - for a photon: ##E=hf=hc/\lambda##
     
  7. Feb 8, 2013 #6

    Dick

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    You were basically right the first time. E=p/c. Just put units on it! And 140eV is not 2.24*10^-34 J.
     
  8. Feb 8, 2013 #7
    Ha, yeah. Its getting late. Its 2.24*10^-17
     
  9. Feb 8, 2013 #8
    ah. haha. thank you mr planck.

    intuitively, i know that for photons, m=0
    but the course doesn't teach it that way. i don't really know what's what.

    E=hf
    f=E/h
    f=3.378582202*10^16 Hz

    λ=v/f
    λ=c/f
    λ=0.000000009m
    λ=9nm

    p=h/λ
    p=7.366666667*10^-26 Js/m

    I don't feel right about the unit.

    h is in Js and lambda is in m
     
  10. Feb 8, 2013 #9

    Dick

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    If you put all that together you just get E=pc again. J*s/m=(kg*m^2/s^2)*(s/m)=kg*m/s. That's a mass times a velocity, so sure it's a good momentum unit.
     
  11. Feb 8, 2013 #10
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