Minimum momentum of electron in a hydrogen atom

[fishinsea]

Homework Statement

The energy of an electron in a hydrogen atom is: E = p^2/2m_e - \alpha e^2/r; where p is the momentum, m_e is the electron charge magnitude, and \alpha the coulomb constant. Use the uncertainty principle to estimate the minimum momentum in terms of m_e, a, e, \hbar.

Homework Equations

\Delta p \Delta r = \hbar/2

The Attempt at a Solution

The answer sheet set dE/dp = 0 to find r, and solved for p using the uncertainty principle, but I'm confused why dE/dP would give you minimum momentum, and whether the minimum momentum corresponds to minimum energy. All the questions I've seen that are related to this confines the electron to a certain radius. If r \to \infty, shouldn't both potential and kinetic energy (thus momentum) go to 0?

 

[mfb]

The ground state has minimum energy. You can find this by varying the radius or the momentum. Both will lead to a rough estimate of the momentum uncertainty in the ground state.

r -> infinity won't give the ground state.

 

[fishinsea]

But why would minimum energy imply minimum momentum? E = KE + PE, if you increase r, the potential energy increases (becomes less negative), allowing for a lower kinetic energy. I'm also tempted to think of this as a gravity/orbit problem, where the velocity would decreases the further away you move away from the rotation center.

 

[mfb]

It does not give the minimal momentum. But you can use it to calculate the minimal momentum uncertainty in the ground state. The momentum uncertainty is smaller for higher energy levels.

 

[fishinsea]

So the momentum uncertainty (and thus momentum) does not occur at ground state? As r -> infinity, would the momentum -> 0?

 

[mfb]

So the momentum uncertainty (and thus momentum) does not occur at ground state?

You mean the minimal? Then yes, otherwise no.

As r -> infinity, would the momentum -> 0?

Yes.

 

[fishinsea]

Okay that cleared up the question mostly, thank you!