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Archived Wavelength for an electron

  1. Sep 25, 2012 #1
    1. The problem statement, all variables and given/known data



    2. Relevant equations

    de Broglie wavelength λ = h/p
    E^2 = p^2 c^2 + (m0)^2 c^4
    L-compton = (h-bar) / mc


    3. The attempt at a solution

    I'm trying to work out the wavelength of an electron with a kinetic energy of 1keV

    So I intend on using
    de Broglie wavelength λ = h/p

    and

    Relativistic mass equation
    E^2 = p^2 c^2 + (m0)^2 c^4

    then inputting the rest mass energy of an electron (511 keV) into the formula above. Though do I need to convert this first to kg?

    Also, I'd still be left with two unknowns in mass eqn i.e. E and p, so how do I obtain the value for E?

    Or should I instead be using the Compton wavelength formula i.e.

    L-compton = (h-bar) / mc

    The question just asks for the wavelength.
     
  2. jcsd
  3. Mar 6, 2017 #2

    QuantumQuest

    User Avatar
    Gold Member

    The relationship between kinetic energy - momentum for a free electron is ##T = \frac{p^{2}}{2m}##. So, from this, we can find the momentum ##p##:

    ##p = \sqrt{2mT} = \sqrt{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19} \times 10^{3}\frac{kg^{2} m^{2}}{s^{2}}} \approx 17.064 \times 10^{-24} \frac{kg m}{s} ##

    Now, we can find the De Broglie wavelength: ##\lambda = \frac{h}{p} = \frac{6.625 \times 10^{-34}}{17.064 \times 10^{-24}} m\approx 0.039 nm##
     
    Last edited: Mar 6, 2017
  4. Mar 23, 2017 #3
    This result is correct, because 1keV is non-relativistic for electrons, since electron rest mass is around 500keV.

    On the other hand one can solve this exercise using the equation OP wanted to use: E² = m²c⁴ + p²c²

    He said the energy (meaning the kinetic energy) of the electron is one keV. Thus 1keV = E - mc² = ##\sqrt{m²c⁴ + p²c²} -mc²##. If OP solves this equation, then OP can insert it in the De Broglie wavelength formula:

    $$p² = \frac{1}{c²}(1keV + mc²)² - m²c² => p = \sqrt{ \frac{1}{c²}(1keV + mc²)² - m²c² } \approx 1.70933440*10^{-23}kg \frac{m}{s}$$

    And so ##\lambda = h/p \approx 0.038764033nm##

    Both results, of course, agree

    EDIT: Not trying to be pedant, but because I want to emphasize the minute differences between both methods I will recalculate what QuantumQuest already did, but with higher precision:

    non-relativistic:

    ##p \approx 1.70849875*10^{-23}kg \frac{m}{s} \approx 31.9687keV/c##
    ##\lambda = h/p \approx 0.038782993nm##

    relativistic:

    ##p \approx 1.70933440*10^{-23}kg \frac{m}{s} \approx 31.98434keV/c##
    ##\lambda = h/p \approx 0.038764033nm##
     
    Last edited: Mar 23, 2017
  5. Mar 23, 2017 #4
    Making my point about the minuteness of the corrections more thorough:

    ##p_{rel} = \sqrt{ \frac{1}{c²}(T + mc²)² - m²c² } = \sqrt{T²/c²+2Tm+m²c²-m²c²} = \sqrt{T²/c²+2Tm} = \sqrt{T}\sqrt{T/c²+2m}##
    ##p_{non-rel} = \sqrt{2Tm} = \sqrt{T}\sqrt{2m}##

    Thus ##\frac{p_{rel}}{p_{non-rel}} = \sqrt{ frac{T/c² + 2m}{2m}} = \sqrt{\frac{T}{2mc²} + 1}##
    And so ##\frac{p_{relativistic}}{p_{non-relativistic}}(x) = \sqrt{\frac{x}{2}+1}## where ##x = \frac{T}{mc²}##

    This formula holds true for any massive particle.

    In our case ##x \approx 1/500## and so the above formula is essentially one.
    This is a good way to see when you can use non-relativistic approximations. Basically as long as your kinetic energies are much smaller than your rest mass energies, you are safe:

    $$ T << mc² $$
     
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