# Coulomb barrier in alpha decay

1. Sep 2, 2007

### seerongo

Can you gurus help me understand in a qualitative way the nature of the Coulomb barrier as it applies to alpha decay? I can intuitively appreciate the Coulomb barrier as it applies to an incoming charged particle, but resources I have been reading apply the same term to the barrier felt by alphas, within the nucleus, inhibiting emission. I would have thought that any barrier that must be overcome by an alpha trying to escape would derive from the dominance of the strong force, binding the alpha to the nucleus, over the Coulomb force of the protons trying to push the alpha out. So, what am I missing here?
By the way, thanks for a great forum.

2. Sep 2, 2007

3. Sep 3, 2007

### seerongo

Yeah, that's what I was hoping for. Thanks for the reply.

4. Feb 5, 2010

### joebo

i have the same question,and still cannot understand the concept, what is coulomb barrier? just like what seerongo said, "any barrier that must be overcome by an alpha trying to escape would derive from the dominance of the strong force, binding the alpha to the nucleus, over the Coulomb force of the protons trying to push the alpha out."
thank you .

5. Feb 16, 2010

### joebo

nobody help me ?

6. Feb 16, 2010

### Staff: Mentor

One is more or less correct. The potential energy is a balance between the nuclear potential energy and the Coulomb potential energy. The nuclear forces act over a shorter distance than the Coulombic force.

With respect to the Coulomb barrier, if the alpha particle, or a pair of (n,p) get beyond a point where the nuclear attraction is exceeded by the Coulomb repulsion, then an alpha particle is released - putting it rather crudely.

Now if one is looking at alpha (Rutherford) scattering, scattering is much more probably than say an absorption (fusion) of an alpha particle by a nucleus. One such reaction is $\alpha$ + 9Be => 12C + n. There are others, and one can look at the absorption (reaction) cross-section as compared to the scattering cross-section.

7. Feb 16, 2010

### Useful nucleus

If you calculate the Q-value for alpha-decay for the nuclei that are known to undergo this type of decay, you get a +ve value. This means that it is energetically favorable for the alpha particle to get out of these nuclei. In fact most of this +ve Q value can be regarded as kinetic energy for the alpha particle inside the nucleus.

Now imagine that the alpha particle pre-exists in the nucleus with kinetic energy enough to escape the potential well of the nuclear forces. Once it exceeds the nucleat radius , all of sudden it finds itself in a (+ve) pure coulombic barrier (Recall that nuclear forces are short ranged and they do not exist beyond the nuclear radius). Classically this region (i.e. the one just beyond the nuclear radius) is forbidden , because all of a sudden the alpha particle gained extremely large +ve potential energy (the coulmbic one) which violates energy conservation. This region forms the barrier. It is a reion prohibited classically but can be penetrated quantum mechanically.

If you look at the first link that Astronuc posted, you will notice that this is the region in dark gray.