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Electron and proton collision

  1. Jun 22, 2010 #1
    At a certain distance in vacuum there are proton and electron (they are at rest with resptect to their's CM). I wonder after what time will they collide?
    Of course, we neglect the fact that when the distance between the particles become very small, there is not only an electrical force acting on them.
    Is this just a two-body problem? Then, what is U(r) in
    [tex]L=\mu \dot{\mathbf{r}}^2/2 - U(r)[/tex]
    where [tex]\mu[/tex] is reduced mass?
     
  2. jcsd
  3. Jun 22, 2010 #2
    use the law of conservation of energy to find the speed at any distance between them:

    [tex]
    E = -\frac{k e^{2}}{r_{0}} = \frac{\mu \, v^{2}}{2} - \frac{k e^{2}}{r}
    [/tex]

    where [itex]r_{0}[/itex] is the initial radius of the problem. Solve this equation for v. Then, use the definition of v as the rate of change of the distance. But, since the distance is decreasing, you will have:

    [tex]
    v = -\frac{dr}{dt}
    [/tex]

    Integrate this 1st order ODE with the initial condtion:
    [tex]
    r(t = 0) = r_{0}
    [/tex]

    and the time of fall T is given by the following condition.
    [tex]
    r(t = T) = 0
    [/tex]
     
  4. Jun 22, 2010 #3

    K^2

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    Science Advisor

    This isn't a very useful problem, though. Electron interacts with proton as a wave. Your initial state with electron at rest some distance from proton is technically some crazy excited state of a hydrogen. When will electron in a hydrogen collide with a proton? Never. Though, there could be overlaps of their wave functions.

    But if you just want to solve it like a classical problem of two point-charges with mass while disregarding any real-world applications, Dickfore's method will produce a result. Might be easier to have r(0)=0 and r(T)=r0, but it'd be purely a mathematical shortcut.
     
  5. Jun 23, 2010 #4
    Dickfore,
    Thanks, but the question was about U. I didn't ask how to solve this problem. :) Anyway the approach using the law of conservation of energy leads to 1st order ODE instead of 2 order ODE for equation of motion. This is good.
    K^2 Of course, it is a fiction problem. I should wrote something like two point charges placed some distance apart from each other.
     
  6. Jun 23, 2010 #5

    K^2

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    Science Advisor

    Fair enough. In that case, yes, just use conservation of energy and integrate.
     
  7. Jun 23, 2010 #6
    The potential energy between two charges is:
    [tex]
    U(r) = k_{0} \frac{Q_{1} \, Q_{2}}{r}
    [/tex]

    where [itex]Q_{1}[/itex] and [itex]Q_{2}[/itex] are the quantities of electric charge of each of the charge, [itex]r[/itex] is their mutual distance and [itex]k_{0}[/itex] is the Coulomb's constant and has different values in different systems of units:

    In SI:
    [tex]
    k_{0} = \frac{\mu_{0}}{4 \, \pi} c^{2}
    [/tex]

    In ESCGS:
    [tex]
    k_{0} = 1
    [/tex]
    and in Gaussian system:
    [tex]
    k_{0} = \frac{1}{4 \pi}
    [/tex]

    The charge of the proton is [itex]+e[/itex] and the charge of the electron is [itex]-e[/itex], where [itex]e[/itex] is the elementary electric charge.

    A dimensionless number independent of the system of units is:
    [tex]
    \alpha \equiv \frac{k_{0} e^{2}}{\hbar \, c} = \frac{1}{137.0}
    [/tex]
     
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