Tunneling of alpha particles

Consider this explanation of the alpha decay: It says

The Coulomb barrier faced by an alpha particle with this energy is about 26 MeV, so by classical physics it cannot escape at all. Quantum mechanical tunneling gives a small probability that the alpha can penetrate the barrier.

The explaining sentence and the attached picture suggest that the alpha particle has to overcome the coulomb barrier in order to escape - and as is hasn't enough energy to do that, it must tunnel through it.

But the Coulomb force is a repulsive force in this case as both the nucleus and the alpha particle are positively charged. So why is there any Coulomb barrier at all, shouldn't the Coulomb force "help" the alpha particle to escape? (The other way round, it's obvious - the particle wants to enter the nucleus, but it is hindered by the repulsive force. But when it wants to leave the nucleus, it is only retained by the nuclear force, as far as I can see.)

Why does the alpha particle have to go trough the coulomb barrier - or am I just misinterpreting the explanation and the tunneling occurs somewhere else?

-
This is quantum mechanics. The alpha particle has a probability of going through the barrier. It only "has to" depending on the probability. – anna v Jul 24 '15 at 18:43

This model is something that I've seen a lot of students (including myself) struggle with, and a big part of it is that calling it the "Coulomb barrier" is quite over-simplified.

Going back to fundamental physics principles, the Coulomb force is repulsive as you say, so I'll write this as $F\propto 1/r^2$, where $F$ is the force and $r$ is the distance from the center of the nucleus and the positive sign indicates that the force is in the positive $r$ direction, away from the nucleus. The potential, which is what's shown on the graph, is the negative of the integral of the field (which is proportional to the force), this results in $E\propto 1/r$. For an analog, imagine lifting a weight where "up" is the positive z-axis. The path involves an increase in z-coordinate and the force from the field (gravitational) is negative. The negative integral of the field over distance is the potential.

The missing part is why they "chop off" this $1/r$ function at some radius, presumably the radius of the nucleus. Well, this allows us to infer some things about the other force present, which is the nuclear force. It would appear that the force acts weakly beyond the radius of the nucleus, since the $1/r$ form is unaltered beyond that point. It also absolutely must have a potential that increases faster than $1/r$ as $r \rightarrow 0$. Both of these would be loosely satisfied if the nuclear potential was $-1/r^n$ (note the negative) where $n>1$. The alpha particle, while in the nucleus, by the way, has kinetic energy which is the reason its energy is higher than the hypothetically lowest energy level possible for it. This has to do with the fact that quantum mechanics only allows certain (quantized) energy levels.

Just For an example of how this is possible, I'll say $E_{nuclear}=-1/r^4$ (I am not saying this is how the force acts, it's just for utility).

$$E(r) = E_{nuclear}(r) + E_{Coulomb}(r)$$ $$E(r) = 1/r - 1/r^4$$

$$F(r) = -\frac{d}{dr} \left( E_{nuclear}(r) + E_{Coulomb}(r) \right) = F_{nuclear}(r) + F_{Coulomb}(r)$$ $$F(r) = 1/r^2 - 1/r^5$$

My intent is that this answer allows you to explicitly answer to yourself what the potential and force contributions are, and give some very rudimentary example for how the "edge" can appear. Beyond that I hope it's clear how someone may take the shape of the above curve and lump it into a "wall".

-
Thank you for your detailed explanation! To make sure I have understood it correctly: The main point is that the "edge" of the potential is not solely due to the Coulomb force, it originates from the addition of Coulomb and nuclear force, right? So the force the alpha particle has to overcome actually is the nuclear force. – diabonas Jun 7 '11 at 16:24
IMHO you are 100% correct in your wording here. The Coulomb force points outward from the center at all points. The extremely strong inward force is nuclear. If there were no Coulomb force, there would be nowhere to escape to. But that's a little bit tricky, since if there were no Coulomb force, the $\alpha$ particle in my graph here would not be able to have $E>0$ in the first place. In other words, without the Coulomb force, the nucleus would be like a finger-trap-of-death for the $\alpha$ with no way out. We might be better off calling it the nuclear-coulomb wall. – Alan Rominger Jun 7 '11 at 16:34
Wonderful, thank you very much :-) – diabonas Jun 7 '11 at 16:40

As you state the Coulomb force is repulsive, this is why the potential increases with smaller spearation. But at small enough distances the strong force takes over which has an attracting nature. The strong force is stronger (hence the name) than the Coulomb force, which is why in total the alpha particle in the nucleus sits at a lower potential than the Coulomb force at the edge.

The strong nuclear force can be modeled as a rectangular potential well to first order.

-