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    I've been asked to work out the momentum of a photon, since they have a mass of zero, how would I do this, because I can't work out the de Broglie's wavelength.
    I know p=mv, but m=0, hence mv=0, does this mean p=o too.
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    Use the following equations

    c = f*lambda
    E = hf
    p = h/lambda
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    thanks very much
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    Does the mass of a photon really equal 0, though?
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    (Original post by Mathematician!)
    Does the mass of a photon really equal 0, though?
    Yes. Well zero rest mass.

     E^2=p^2c^2+m^2_0 c^4

    So it still has momentum even without rest mass.
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    (Original post by TableChair)
    Yes. Well zero rest mass.

     E^2=p^2c^2+m^2_0 c^4

    So it still has momentum even without rest mass.
    Don't even need to make it that complicated...its massless...so just E=cp
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    (Original post by -G-a-v-)
    Don't even need to make it that complicated...its massless...so just E=cp
    Energy = speed of light X momentum ?

    Srry, how did u get this equation?

    Thanks.
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    (Original post by -G-a-v-)
    Don't even need to make it that complicated...its massless...so just E=cp
    Yes I know, but I'm showing where it comes from.
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    (Original post by wizz_kid)
    Energy = speed of light X momentum ?

    Srry, how did u get this equation?

    Thanks.
    Two ways, either use the equations in my first post, or set  m_0 = 0 in the the one I gave in my second.
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    (Original post by TableChair)
    Yes. Well zero rest mass.

     E^2=p^2c^2+m^2_0 c^4

    So it still has momentum even without rest mass.
    In fact this is where the famous E = mc^2 comes from; we assume the first term is negligible compared with the second (because of the relative powers of c).
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    (Original post by TableChair)
    Yes. Well zero rest mass.

     E^2=p^2c^2+m^2_0 c^4

    So it still has momentum even without rest mass.
    Oh OK. That is interesting... What does "zero rest mass" actually mean? (What do you mean by rest? Completely stationary?)
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    rest mass is the energy associated with a body/particle that is not due to motion (i.e. the bit left over after you deduct kinetic energy).
 
 
 
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