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

[geophysics10]

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?

 

[Simon Bridge]

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 $q_e$.

 

[geophysics10]

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!

 

[Simon Bridge]

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.