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Enclosed charge of an electron cloud with a given charge density function.

  1. Sep 24, 2012 #1
    I have a conceptual problem.

    1. The problem statement, all variables and given/known data
    I was given a charge distribution for an electron cloud of a hydrogen atom in the ground state - ignoring the nucleus.


    2. Relevant equations
    charge density: ρ=charge of electron/(pi*Bohr radius^3)*exp(-2r/Bohr radius)


    3. The attempt at a solution
    Whenever I integrate over the volume of the sphere to find the enclosed charge, I got a value other than the charge of an electron. Why isn't the enclosed charge equal to the charge of an electron? What concept am I missing here? The charge density function comes from quantum mechanics. Is there some weird quantum effect there that is taken into account?
     
  2. jcsd
  3. Sep 24, 2012 #2

    Simon Bridge

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    The charge distribution does not terminate inside a finite volume - plot ##\rho(r)## vs ##r## and compare with the radius of the volume you are integrating over.

    The "weird quantum effect" is the statistical nature of that charge distribution ... for a sphere radius R (finite volume) about the nucleus, there is always a non-zero probability of finding the electron outside that volume. Integrate to infinity and you should get ##q_e##.
     
  4. Sep 25, 2012 #3
    Thank you. I took a drive and realized that some of the distribution must lie outside the Bohr radius because it is an average value for the distance of the electron, and that when r is at infinity the electron and proton must both be point charges at the origin. I'm glad I came back to someone confirming that!
     
  5. Sep 25, 2012 #4

    Simon Bridge

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    Yah - this treatment of the electron as having its charge smeared out over the wave0-function is only an average treatment. An actual interaction may measure the position of a particular electron more accurately than that... but, for a large number of hydrogen atoms, it all averages out.
     
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