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De Broglie wavelength (relativistic e-)

  1. Jun 9, 2009 #1
    Wavelength of an electron

    1. The problem statement, all variables and given/known data

    -Electron has 3.00 MeV (or 4.8*10^-13 Joules)
    -it's relativistic
    -finding λ.

    2. Relevant equations

    h=6.63*10^-34

    λ=h/p (obviously)

    And I'm not sure if they're needed, but the relativistic eq's are:

    KE = mc^2/sqrt(1-(v/c)^2)
    p = mv/sqrt(1-(v/c)^2)

    I'm not sure if this one applies to relativistic speeds:

    E = hc/λ

    3. The attempt at a solution

    Attempt 1:

    E = hc/λ

    4.8E-13 = (6.63E-34)(3E8)/λ
    λ = (6.63E-34)(3E8)/(4.8E-13)
    λ = 4.14E-13 m

    BUT answer key says 3.58E-13

    If you could help, that would be great.
    Sorry if it's too long, and I'm a little unfamiliar with relativistic eqn's so forgive me if I screwed up on them.
     
    Last edited: Jun 9, 2009
  2. jcsd
  3. Jun 9, 2009 #2

    Redbelly98

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    Re: Wavelength of an electron

    Actually, this "KE" expresson is giving the total energy, kinetic + rest mass energy, so

    KE + mc2 = mc2/sqrt(1-(v/c)2)

    That's an approximation that applies at extremely relativistic speeds (say v>0.99c), and is strictly true only when v=c, i.e. for photons and other massless particles.

    Since this is a moderately relativistic situation, E=hc/λ is not valid.

    You could try using the KE + mc2 equation instead, but many problems like this one make use of this:
    E2 = (mc2)2 + (pc)2
    where, again, E is the total energy,
    E = KE + mc2
     
  4. Jun 9, 2009 #3

    Redbelly98

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    Re: Wavelength of an electron

    Wait, I just realized this is close to an extreme relativistic situation.

    Yes, that will work. However, E is the total energy, kinetic + rest mass energy. Just using the kinetic energy for E is wrong.
     
  5. Jun 9, 2009 #4
    oookay that makes a lot more sense. Thanks so much, redbelly. :)
     
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