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Homework Help: Question regarding the de Broglie wavelength of a particle.

  1. Feb 7, 2014 #1
    1. The problem statement, all variables and given/known data

    A particle with charge e and mass m_0 is accelerated by a potential V to a relativistic speed.

    Show that the de Broglie wavelength is:

    λ = h / √(2 m_0 eV ) √(1+( eV / 2 m_0 c^2 ))

    2. Relevant equations
    E = qV
    λ = h/p
    E = √(p^2 * c^2 + m^2 * c^4 )
    p = (m_0 * v) / (1 - (v^2/c^2)

    3. The attempt at a solution
    We do not have an expression for v, and so we use the formula:
    E = √(p^2 * c^2 + m^2 * c^4 )
    which gives us that:
    E^2/c^2 = p^2 +m^2 *e^2
    Which gives:
    p = √(E^2/c^2 + m^2 * c^2)
    Putting p into the expression for the de Broglie wavelenght gives us:

    λ = h/ √(E^2 / c^2 + m^2*c^2)

    We know that the potential energy of the particle in the Field before accelleration starts (and therefore the kinetic energy) is simply eV, and put that into the expression as well.

    λ = h/ √(eV^2 / c^2 + m^2*c^2)

    It has been a very long time since I last did any real maths, super basic algebra was sufficient for most of what I've done for the last two years. I am guessing the path to making the expressions the same lies in derivation, though why that should be an operation I have no idea.

    I am well and truly lost on this, so if anyone have any insight whatsoever, I will be very grateful.
    (Oh, and this is one of several attempts down different roads. I think this is the one that is closest to a proper solution, and also the one that is the least confusing for those of you who actually know this stuff...)
    Last edited: Feb 8, 2014
  2. jcsd
  3. Feb 7, 2014 #2


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    The kinetic energy of the particle is eV. The total energy E is the sum of the kinetic and rest energies. If you fix that, you'll probably see how to get it to work out.
  4. Feb 8, 2014 #3
    Thank you so much for your reply, I'm still not entirely sure about how to og about this. I think the formula used for the energy in my post above is the formula for the kinetic energy, and the formula for the total energy would be:
    Ek + mc2 = √(p^2 * c^2 + m^2*c^4)

    which turns into:

    Ek = √(p^2 * c^2 + m^2*c^4) - mc2
    Ek 2 = p2 * c2 + 2m2*c4
    Ek 2 / c2 = p2 +2m2 * c2
    p2 = Ek 2 / c2 - 2m2 * c2

    p = Ek / c - √(2m2 * c2)
    p = √(eV2 / c2 -2m2 c2)

    λ = h / √(eV2 / c2 -2m2 c2)

    Is there a difference in the two masses? Should I put in a Lorentz factor somewhere to account for rest mass vs. mass at relativistic speed?
    Last edited: Feb 8, 2014
  5. Feb 8, 2014 #4


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    I think, here we have to give some general concepts, before helping to solve the problem. It is very important to get the right idea about the quantities involved in relativistic particle physics, particularly not to confuse the students with unnecessary old ideas about "relativistic mass" etc. (see Simen's previous posting concerning his confusion/uncertainty about possible Lorentz factors!).

    The de Broglie wavelength for a particle with momentum [itex]p[/itex] is given via the wave number of the corresponding wave mode in quantum field theory, i.e.,
    [tex]k=\frac{p}{\hbar}=\frac{2 \pi}{\lambda} \; \Rightarrow \; \lambda=\frac{2 \pi \hbar}{p}=\frac{h}{p}.[/tex]

    For the given question, the total energy of the particle is given by
    [tex]E=m_0 c^2 + |e V|=\sqrt{m_0^2 c^4 + p^2 c^2}, \qquad (1)[/tex]
    because the kinetic energy gained when a charged particle is accelerated starting at rest running through a electrostatic potential difference [itex]V[/itex] is given by [itex]|e V|[/itex], and in relativistic physics we always include the "rest energy" [itex]m_0c^2[/itex] in the total energy of the particle. That's very convenient, because then energy [itex]E[/itex] and momentum [itex]\vec{p}[/itex] together build a Minkowski four-vector which is manifestly Lorentz covariant. The invariant square of this four-vector is invariant under Lorentz transformations and given by
    [tex]E^2/c^2-\vec{p}^2=m_0^2 c^2.[/tex]
    This shows that the mass of a particle is a scalar under Lorentz transformations and thus independent of the particle's velocity.

    It's a very good practice in modern relativistic particle physics that we only use the notion of mass in the sense of "invariant mass", which is a Lorentz scalar. For a massive particle the invariant mass is sometimes also called the "rest mass" of the particle. For a massless particle, the invariant mass is zero, and the particle always moves with the speed of light.

    For your problem you solve (1) for [itex]p[/itex] as a function of the invariant mass [itex]m_0[/itex] and the kinetic energy [itex]|e V|[/itex]. This should lead you to the correct formula for the de Broglie wavelength.

    If you are supposed to give the result in terms of the (three-)velocity of the particle, then you can use
    [tex]p=\frac{m_0 v}{\sqrt{1-v^2/c^2}}.[/tex]
    It's easy to show that the energy of the particle (inlcuding its rest energy) is given by
    [tex]E=\frac{m_0 c^2}{\sqrt{1-v^2/c^2}}.[/tex]

    Note: In the old days, before Minkowski had clarified the mathematical structure of relativistic spacetime (Minkowski space), the people used to introduce several kinds of "relativistic masses" that are speed dependent. Although unfortunately some modern textbooks still use these old-fashioned ideas, this is more confusing than helpful. So it's better to work only with the invariant mass, energy, and momentum.
  6. Feb 8, 2014 #5


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    As vanhees noted, you should expunge the notion of relativistic mass from your brain. You also made some algebra errors. You seemed to have used ##(a+b)^2 = a^2+b^2## when squaring the expression for ##E_k## and ##\sqrt{a^2+b^2}=a+b## when solving for ##p##. Neither is correct.
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