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Mass in eV over speed of light

  1. Apr 11, 2015 #1
    I know the mass of, for instance, an electron can be express in units of electronvolts over speed of light squared. However, I am not sure how to use it. When I have to calculate [itex]\frac{\sqrt{2mE}}{\hbar}[/itex], do I have to multiply m with the speed of light squared to get it in meters per second squared? Or, divide? Yes, I am a little confused.

    PS. Energy is given in electronvolts, and Planck's reduced constant is given in electronvolt seconds.
  2. jcsd
  3. Apr 11, 2015 #2
    Okay, but, E is in electronvolts. Also, I have other constants that I need to calculate as well. Would you recommend me to convert all potentials and energies to joules?

    Lets say that is not possible. It is exam, and I have to write a code on paper that works (which does happen here).
  4. Apr 11, 2015 #3


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    Starting with E = mc2, then m = E/c2, and one can express E in eV, or MeV, and divide it by c2.

    So mE leads to (E/c2), or (E/c)2.

    Note k = p/ħ

    Units should be consistent.
  5. Apr 11, 2015 #4
    Can I do that? I mean, isn't that only valid if my particle is not moving? My particle is moving. I think the mass I am using is the rest mass, but the energy certainly is not that of a still particle.

    I should probably provide some context; I encounter this issue whenever I have to program some solution to the Schrodinger equation.
  6. Apr 12, 2015 #5
    Schroedinger is non-relativistic so you can: you're assuming the gamma factor is very close to 1, so the eV mass unit is to be seen as a natural units convention instead of as a full-fledged relativistic approach. Read a booklet on natural units if you need to. If you tried to introduce the gamma factor you'd have to use the Dirac equation to be consistent, but then the rest mass is the one appearing in the Dirac equation.
  7. Apr 13, 2015 #6


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    First of all that's indeed a non-relativistic energy-momentum relation, i.e., ##E## should be the kinetic energy of your particle. Then ##p=\sqrt{2mE}## is the magnitude of the particle's momentum and ##k=p/\hbar## is the wave number and ##\lambda=2\pi/k## the de Broglie wavelength. The mass in modern physics is always the invariant mass of the particle. The socalled relativistic mass shouldn't be used anymore anyway, but that's another story.

    Further, if written in correct equations, physics is independent of the system of units used (one has to be careful in electrodynamics, where various different systems of units are in use: SI (very unpleasant for theoretical physics but common in experimental physics and because it's the official system of units), Gaussian and Heaviside-Lorentz units, which differ only in the appearance or non-appearance of factors ##4 \pi## in Maxwell's equations).

    The usual system of units in high-energy physics is to use MeV or GeV for particle masses (then giving ##E_0=m c^2## of course) and fm for lengths and fm/c for times, where fm=femto metres=fermi=##10^{-15} \; \mathrm{m}##.

    It's most convenient to set ##\hbar=c=1## ("natural units"). The only quantity you need to remember then is ##\hbar c \simeq 0.197 \; \mathrm{GeV} \mathrm{fm} = 197 \; \mathrm{MeV} \mathrm{fm}##. In this units it's very simple to work with, because masses, energies and momenta all have the same units. Usually one uses MeV or GeV.

    For your example: Say, we look at an electron. It's mass in natural units is ##m=0.511 \; \mathrm{MeV}##. Now let's look at a non-relativistic electron with a kinetic energy ##E=10 \; \mathrm{keV}=10^{-2} \; \mathrm{MeV}##. Then the momentum is
    $$p=\sqrt{2 m E}=\sqrt{2\cdot 0.511 \cdot 10^{-2}} \; \mathrm{MeV} \simeq 0.101 \; \mathrm{MeV}.$$
    The wavenumber then is given by
    $$k=0.101 \mathrm{MeV}/(0.197 \; \mathrm{MeV} \mathrm{fm}) \simeq 0.513/\mathrm{fm}.$$
    Last edited: Apr 14, 2015
  8. Apr 13, 2015 #7
    Okay, I think I get it. Thank you all for your answers!
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