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

  • #1
I have a conceptual problem.

Homework Statement


I was given a charge distribution for an electron cloud of a hydrogen atom in the ground state - ignoring the nucleus.


Homework Equations


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


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?
 

Answers and Replies

  • #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##.
 
  • #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!
 
  • #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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