How can the momentum-position Uncertainty Principle not violate conservation of energy? [duplicate]

As far as I understand, the primary reason why an electron doesn't fall into a nucleus is that, when it gets close, it has a sufficiently high probability of having a lot of momentum to get back to its original position. This high probability, in my understanding, arises because the electron is confined to only a small amount of space around the nucleus, and hence, by Heisenberg's Uncertainty Principle,the uncertainty in its momentum is high, which in turn increases the probability of the momentum being large.

In such cases, where the electron "bounces off" or "phases through" the nucleus, the momentum would always have a non-zero magnitude, and hence the kinetic energy would always be non-zero. This way, energy would be gained with each new "bounce". Additionally, the electrons would be constantly accelerating, radiating away light of all frequencies.

Is my premise incorrect, or is it the implication? And in what way is it incorrect?

• There is another form of the Uncertainty Principle, namely Time-Energy: $\Delta t \Delta E \ge \frac{\hbar}{2}$. Conservation of energy is only true within a timescale $t > \Delta t$; else virtual particles would also violate energy conservation. – Sparrow Jun 22 '19 at 1:20
• Possible duplicate of Energy conservation limited by uncertainty principle – Aaron Stevens Jun 22 '19 at 2:40
• @Aaron Stevens Totally different questions. I have looked at this question before posting this one. My question is about the momentum-position uncertainty; the question you linked is about the energy-time uncertainty. Even if these are identical, I do not understand how, which is the point of this question. – Max Jun 22 '19 at 12:25
• @Sparrow what about the expectation of the electron's energy over a long period of time? Clearly, if the momentum keeps bumping up when it gets close to the nucleus, the expectation value of energy at a given time will grow? – Max Jun 22 '19 at 12:32