What Keeps Electrons Away from the Nucleus of an Atom?

[vorlon13]

Neighborhood redlining, IOWs . . . .

[Banned]

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So there is a playoff between the electromagnetic force and the uncertainty principle making it so even the electrons at the lowest energy have to stay away from the nucleus.

Interesting conclusion.

[polymath257]

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So there is a playoff between the electromagnetic force and the uncertainty principle making it so even the electrons at the lowest energy have to stay away from the nucleus.

Interesting conclusion.

It's pretty standard. Look in any QM physics text.

[shadow]

In fact, the lowest 'orbital' in an atom has no angular momentum! What actually happens is that the Heisenberg uncertainty principle comes into play. Essentially, if the electron tried to collapse close to the nucleus, we would know where it is located, which means its momentum would be uncertain, meaning it wouldn't *stay* close to the nucleus.

I don't understand how the bolded follows from Heisenberg's uncertainty. If I recall my chemistry correctly (probably not) isn't that just a limit of us measuring a particle's velocity or location, in that we can only measure one or the other? So how does that impact the location of the particle itself?

[polymath257]

In fact, the lowest 'orbital' in an atom has no angular momentum! What actually happens is that the Heisenberg uncertainty principle comes into play. Essentially, if the electron tried to collapse close to the nucleus, we would know where it is locatedpermittivity of free space., which means its momentum would be uncertain, meaning it wouldn't *stay* close to the nucleus.

I don't understand how the bolded follows from Heisenberg's uncertainty. If I recall my chemistry correctly (probably not) isn't that just a limit of us measuring a particle's velocity or location, in that we can only measure one or the other? So how does that impact the location of the particle itself?

Think of it like this. If the electron stays too close to the nucleus, that means the position is well determined. By the uncertainty principle, that means the momentum is very undetermined. But for an orbit, that would push the electron either closer or farther away from the nucleus, most likely farther away.In fact, I've seen a derivation of the size of a hydrogen atom using only these properties.

The derivation goes as follows:

The potential energy for an electron at a distance x from an inverse square force (like the Coulomb law) is Q=-e^2/(4*pi*eps*x). Here, e is the charge of the electron and eps is the  permittivity of free space.

Now, since we are only doing first approximations, we can estimate the uncertainty of position by x. The corresponding uncertainty in momentum is p=hbar/x where hbar is Planck's constant divided by 2pi.

This gives a kinetic energy of p^2/(2m) where m is the mass of the electron.

Now, for an orbit, the potential and kinetic energies have to balance, so we get e^2/(4*pi*eps*x)=hbar^2/(2*m*x^2), which gives

x=hbar^2*4*pi*eps/(2*m*e^e).

This turns out to give exactly the Bohr radius of the hydrogen atom (even though we played fast and loose with the constants).