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Relativistic Wavelength of Electron in Transmission Electron Microscop

  1. Feb 6, 2014 #1
    [I could not fit the last "e" into the title]


    I am writing a report on TEM, and was asked to consider the effect of accelerating voltage on the resolution. Since resolution in the light microscope is limited by the wavelength of visible light, we obtain much higher resolutions in a TEM due to the wave-particle duality of matter that de Broglie found.

    So, I found a relativistic expression of the electron wavelength online, and it goes like this:

    \lambda = \sqrt{\frac{h^2 c^2}{eV \cdot [2m_0c^2 + eV]}}

    I wanted to do a little derivation of this expression. I started from the relativistic energy of a particle as described by Einstein in his special relativity:

    E = \sqrt{p^2 c^2 + m_{0}^{2} c^4}

    I then substituted [itex] p = \frac{h}{\lambda} [/itex] into the equation, and obtained:

    E = \sqrt{\frac{h^2}{\lambda^2}c^2 + m_{0}^{2} c^4}

    Solving for [itex]\lambda[/itex], I got:

    \lambda = \sqrt{\frac{h^2 c^2}{E^2-m_{0}^{2} c^4}}, equation 1

    Next, I use that the kinetic energy of an electron with charge e in a potential V is expressed as:

    eV = \frac{1}{2}m_0 v^2

    Multiplying both sides with 2m and taking the root, I obtain

    mv = \sqrt{2m_oeV} = p

    Substituting the above for [itex] E = pc [/itex] and squaring yields

    E^2 = c^2 2m_o eV

    Now I plug this into equation 1, which ultimately gives me

    \lambda = \sqrt{\frac{h^2}{m_0[2eV-m_0 c^2]}}

    Now, this expression is similar, but not identical to the one I stated at the beginning. Where am I going wrong. Or what have I forgotten to do? Could someone help me with this? This derivation is way out of the scope of my TEM exercise, so it is purely out of curiosity. Am I even going in the correct direction with this derivation?

    Kind regards,
    Last edited: Feb 6, 2014
  2. jcsd
  3. Feb 6, 2014 #2


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    Staff: Mentor

    This is correct, so far.

    E is the total energy of the electron (rest-energy plus kinetic energy). V is the potential difference through which the electron is accelerated in the TEM.

    When you accelerate an electron with charge q, from rest through a potential difference V, it ends up with kinetic energy equal to qV.

    This should give you enough information to complete the derivation.

    Also, your first equation (the one you found somewhere) is incorrect. The units are inconsistent. In the denominator, m0c2 has units of energy (joules), and V has units of volts. You can't add them.

    Nevertheless, that equation is close to the correct one. If you finish your derivation correctly, you'll see what's missing. There's a clue in my statement about accelerating the electron, three paragraphs above this one. :smile:
    Last edited: Feb 6, 2014
  4. Feb 6, 2014 #3


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    Staff: Mentor

    Ah, you were working while I was typing...

    That doesn't work for electrons, only for particles with m0 = 0 (e.g. photons).

    You shouldn't need to use the velocity v at all in this derivation.
  5. Feb 6, 2014 #4
    Thank you! I updated the post just before you answered. But I not sure I have made it. My expression leaves me with a complex wavelength.

    edit: Man, it happened again!
  6. Feb 6, 2014 #5
    Ah, yes, E = pc is a special case for the einstein relativistic energy. I will give it another try.
  7. Feb 6, 2014 #6
    Yes, I did it! Starting from scratch with equation 1, which was

    \lambda = \sqrt{\frac{h^2c^2}{E^2 - m_{0}^2c^4}}

    You said E represents the total energy, the sum of the kinetic energy and the rest energy.

    The rest energy is given as the equation already stated at the beginning (relativistic energy of particle). The kinetic energy of a charged particle in a potential is then given as

    E_K = \frac{1}{2}mv^2 = eV, equation 2

    I then say the total energy is

    E_{total} =eV + \sqrt{p^2c^2 + m_{0}^2 c^4}

    When I plug this into equation 1, the two terms [itex] \pm m_0^2c^4 [/itex] cancels. This leaves:

    \lambda = \sqrt{\frac{h^2c^2}{(eV)^2 + p^2c^2}}, equation 3

    Again, solving equation 2 for [itex] (mv)^2 = p^2 [/itex] gives

    p^2 = 2m_0eV

    which I plug into equation 3. This gives, after some cleaning

    \lambda = \sqrt{\frac{h^2c^2}{eV[2m_0c^2 + eV]}}

    Thank you jtbell for your help! It always feels great to derive something when you have spent an hour or more trying! Thanks again!
  8. Feb 6, 2014 #7


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    Staff: Mentor

    But ##\sqrt{p^2c^2 + m_{0}^2 c^4}## all by itself is the total energy! I think you may be getting confused by all the different E's flying around. The E in your equation 1 is the total energy, so it's synonymous with ##E_{total}##. So you have ##E = E_{kinetic} + E_{rest}##.

    ##E_{kinetic} = eV##. Now, what is ##E_{rest}##?

    You did in fact end up with the correct equation, so there's another mistake somewhere along the line that "un-did" your first mistake.
  9. Feb 8, 2014 #8
    Oh. I thought in the lines of that since the relativistic energy is expressed in terms of the rest mass, that equaled the rest energy, i.e. the energy of the electron at rest.

    If I just substitute [itex]E^2[/itex] with [itex]E = \sqrt{p^2c^2+m_0^2c^4}[/itex], I end up with:

    \lambda = \sqrt{\frac{h^2c^2}{p^2c^2}}

    The [itex]c[/itex]'s cancel, and I can plug in [itex] p^2 = m_0^2v^2 = 2m_0eV [/itex] to obtain

    \lambda = \sqrt{\frac{h^2}{2m_0eV}}

    which is far from the correct expression.

    But then again, [itex]E = \sqrt{p^2c^2+m_0^2c^4}[/itex] includes the momentum, and so I agree that would be the total energy. I am not sure what to do now, honestly.
  10. Feb 8, 2014 #9


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    OK... your starting point is
    $$\lambda = \sqrt{\frac{h^2c^2}{E^2 - m_{0}^2c^4}}$$
    E is the total energy, namely the sum of the rest and kinetic energies, so
    $$\lambda = \sqrt{\frac{h^2c^2}{(E_{rest} + E_{kinetic})^2 - m_{0}^2c^4}}$$
    You know the kinetic energy is eV. What is the rest-energy? Substitute them and simplify...
  11. Feb 17, 2017 #10
    I have a question about this formula.

    If Ekinetic = 0, we have:

    E total= Erest

    This term is squared and then you substract the same term which is m02 c4.

    Don't you get a zero on the denominator? Which doesn't make sense. Well I guess it does make sense, it means there is not any wavelenght associated with a particle if it is at rest right? I mean it is infinite, so there is no oscillatori behavior associated with it.

    And I have a second question.

    When you are given the Kinetic Energy in eV you can add to it the rest energy to get the total energy.

    Does that mean that the kinetic energy is equal to p2 * c2 ?

    From relativity Energy:

    E2 = p2c2 + m2c4
  12. Feb 17, 2017 #11
    I add another question on the same topic.

    What is the wavelenght of an electron with Kinetic Energy 10 ev?

    If you use Classical Mechanics you get:

    λ=h/ √2 m E

    m=9.1 x 10 -31

    E=10 ev * 1,6 x 10 -19 = 1.6 10 -18 Joule

    h= 6.63 10-34

    So I get for λ:

    3.88 10-10 meters

    But if I use relativity, I can calculate λ using the relativity equation for the Energy so:

    λ = hc/√(Etotal)2 -(Erest)2

    I get 1.24 10-7 meters

    This is around 1000 times bigger than the Classical value.

    So, I guess the relativistic value is allways correct, and the classical value just an approximation, Is this correct?

    In this case it looks a bad approximation.
  13. Feb 17, 2017 #12


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    Staff: Mentor

    No. If it did, then:

    would give ##E^2 = E_{kinetic} + m^2 c^4 = E_{kinetic} + E_{rest}^2## which isn't correct. The units don't match up!
    Last edited: Feb 17, 2017
  14. Feb 17, 2017 #13


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    Staff: Mentor

    This is OK.

    Show your arithmetic. The result should be very very close to your previous value. In general, the classical approximation (your first equation) works well if ##E_{kinetic} \ll E_{rest}##. For an electron, ##E_{rest} = 511~\rm{keV} = 511000~\rm{eV}##. Clearly ##10~\rm{eV} \ll 511000~\rm{eV}##.
  15. Feb 18, 2017 #14
    I got it, I missed one square root in one of the steps.

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