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I Electron motion in the plum pudding model

  1. Jun 11, 2016 #1
    Hi, Im trying to work out how the electron would oscillate about a mean position in the plum pudding model.

    Plum pudding model;
    -1 electron atom.
    -Positive charge of ##+e## distributed evenly about the volume of the atom of radius ##R##.
    -Electron (charge ##-e##) is free to move within the sphere.

    My first attempt at this is to take it as a 1D problem with a positive line charge of density ##l=\frac{e}{2R}##. When perturbing the electron along the left side of this line of displacement ##-r##, where the origin is at the centre of the sphere, I evaluated the force on the electron from the two sides to be,

    $$F_L=-\frac{e^2}{8\pi \epsilon_0 R}\int_{-r}^{-R}r^{-2}dr=-\alpha\left(R^{-1}-r^{-1}\right)$$.

    With ##\alpha=\frac{e^2}{8\pi \epsilon_0 R}##, ##F_L## is the force from the left side charge distribution. And for the right side,

    $$F_R=-\frac{e^2}{8\pi \epsilon_0 R}\int_{-r}^{R}r^{-2}dr=\alpha\left(R^{-1}+r^{-1}\right)$$.

    Where for both I have taken infinitesimal force ##dF##, to be,

    $$dF=-\frac{edq}{4\pi \epsilon_0 r^2}$$.

    Where ##dq=ldr##. So the net force on the electron (##F_L+F_R##) will be able to give us the equation of motion for the electron, which is

    $$\ddot{r}-\frac{\alpha}{m}\frac{1}{r}=0$$.

    I'm not sure why but I feel like this is wrong, I would expect the solution to the EOM to be simple harmonic but this doesn't seem to give that. Can anyone clarify whether this is an issue with the 1D approximation, or with the workings? Thanks.
     
  2. jcsd
  3. Jun 11, 2016 #2

    Simon Bridge

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    Um... why?
     
  4. Jun 11, 2016 #3
    Why not? It's good practice to see where a model breaks down. To what degree does the oscillation of the electron and its emitted radiation from the acceleration about the equilibrium position agree with measured values. Again, it's good practice to know the things that are wrong and why, rather than just whats correct.
     
  5. Jun 11, 2016 #4

    jtbell

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    Are you familiar with Gauss's Law? If you assume the "pudding" is a sphere with uniform (positive) charge density, you can use Gauss's Law to show that the electric field inside the sphere, a distance r from the center, is proportional to r. The direction of the field is away from the center. Therefore the force on an electron at distance r from the center is also proportional to r, and directed towards the center.
     
  6. Jun 12, 2016 #5
    Ahh, didn't even think to use Gauss's law, have got down to an EOM of what I expected

    $$\ddot{r}+cr=0$$

    Thanks
     
  7. Jun 12, 2016 #6

    Simon Bridge

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    There were a lot of ways to respond. In order to best answer your questions, it is useful to know what your intended purpose/motivation is.

    You have posted a radial eom, orbital motion forms part of Thompson's model too.
    https://en.m.wikipedia.org/wiki/Plum_pudding_model

    Note: it is usually more useful to examine limitations to a model with such a serious flaw (wildly incorrect charge distribution) if its impact of contemporary measuremrnts is examined. ie it was used in early attempts to determine the bohr radius for hydrogen.
     
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