# Minimum momentum of electron in a hydrogen atom

## 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
Mentor
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.

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
Mentor
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.

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

mfb
Mentor
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.

Okay that cleared up the question mostly, thank you!