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Alpha decay part II

Dear Reader,

As I sit typing in a railway carriage on the way home sitting near a young lady who is sporting a ”Nuclear power no thanks” badge, I sit here thinking about nuclear processes hopeful that the young lady does not notice what I am typing.

It is interesting to note that one of the physical effects which regulate the reactions which go on inside the red sun of the “Karnkraft nej tak” badge are the electrostatic forces which oppose fusion. The same forces have an effect on the reverse reactions (alpha emission, fission and all the cluster emissions which come between those two extremes).

Now since I had a rather short hair cut recently I can not demonstrate electrostatic attraction using a comb dragged through my hair. I will let you try that at home, also I do not have a cat to rub on a bit of plastic so I can not use that either.

But back to nuclear processes and electrostatics, to a first approximation the atomic nucleus can be treated as a charged sphere. The size is given by the following equation.

R = Ro (A)0.3333

Where Ro is equal to 1.2 x 10-15 m, while A is the total number of nucleons (the sum of the number of protons and neutrons) in the nucleus.

So radius of a plutonium-238 nucleus is 7.44 fm, while its daughter (uranium-234) has a nuclear radius of 7.39 fm while an alpha particle has a radius of 1.90 fm. Using these radii we can calculate the energy required to push the alpha particles from plutonium-238 back inside the nucleus.

To do this we need a few more equations from A-level physics.

As the capacitance of a sphere is given by

C = 4 π ε r

Where ε is equal to the permittivity of free space which is 8.854187817620 × 10−12 F m−1 or just 8.85 × 10−12 F m−1

We can use this to estimate the electrostatic energy required to hold an alpha particle on the surface of the uranium-234 nucleus. As soon as the alpha particle is taken out of the nucleus it is no longer being strongly bonded by the very short ranged but very strong attraction between the protons and neutrons in the nucleus (the strong force). So suddenly only the electrostatic forces apply to the system (the weak force and gravity are far far weaker)

This energy (28.5 MeV) is far greater than the decay energy of the plutonium-238 (5.593 MeV), as a result the alpha particle, what has to happen is that the alpha particle must overcome this energy barrier before it can leave the nucleus of the atom. What happens is that by quantum tunneling the alpha particle leaves the nucleus of the atom and then goes on its merry way. Here is a graph of the electrostatic energy in MeV vs the distance from the centre of the nucleus for both the alpha particle and the carbon-12 nucleus.

Electrostatic energy as a function of distance from the centre of the daughter nucleus

When the calculation is repeated for the loss of a carbon-12 nucleus from plutonium-238 to form radium-226 then I have estimated that the energy barrier is 74.5 MeV, while the decay energy is now higher at 22.5 MeV you now have a bigger barrier and some other things also help slow down the release of C-12 nuclei.

I hope to get onto these things later.


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