Alpha decay: why Helium nuclei?

[haushofer]

TL;DR Summary

Why do heavy nuclei "prefer" to emit Helium nuclei instead of other nuclei? I don't truly understand the usual "high binding energy for helium nuclei"-argument.

Dear all,

in my teaching of nuclear physics at high school level I noticed that I never really wondered about why alpha decay consists of helium nuclei. So I consulted a lot of lecture notes online, but couldn't find a satisfying answer. The texts I used in the past are "concepts of modern physics" of Beiser, and "Particles and nuclei" of Povh et.al.

Lets focus on the explanation by Beiser, (the Wikipedia page https://en.wikipedia.org/wiki/Alpha_decay , under "Mechanism" is also based on this text). In section 12.4 it says

"Why are alpha particles emitted rather than, say, individual protons or $^3_2He$? The answer follows from the high binding energy of the alpha particle. To escape from the a nucleus, a particle must have kinetic energy, and only the alpha particle mass is sufficiently smaller than that of its constituent nucleons for such energy to be available. "

Beiser then illustrates this by giving the disintegration energy of particle emission by a heavy nucleus,

$$Q = (m_i - m_f - m_x)c^2$$

where "i" is the initial (parent) nucleus, "f" the final nucleus (daughter) and "x" the emitted particle mass. But apparently I'm missing something basic here. Can someone point to a reference where this is worked out in detail, or give babysteps to complete this argument (given the high binding energy per nucleon for helium nuclei; I do understand that)? Apparently, helium nuclei have a small mass (compared to their constituent masses) due to their high binding, which gives a Q which is positive, and this is not the case for other nuclei.

I also tried to describe particle emission via the tunneling mechanism, replacing the standard derivation of alpha decay as helium nuclei by general nuclei, but I couldn't conclude from this why helium nuclei emission is so much more probable than, say, carbon, hydrogen or other light nuclei.

As I said, I think I miss something really basic here. Online I couldn't find much more explanation than "alpha particles have high binding energy per nucleon", so apparently it's quite easy to understand. Any input is appreciated :)

 

[fresh_42]

My book (Mayer-Kuckuk) explains it this way:
(ref. Hanna, G.C. in: Experimental Nuclear Physics, ed. E. Segrè, Vol. III NY 1959)

• Calculate the resulting (available) kinetic energy in dependence of the masses involved.
• Regard the diagramm $\text{ # Neutrons} / \text{ # Protons} = f(\text{ # Nuclei})$ and draw in the limits of stabilities computed by Hanna's method.
• Observe that the energy isobars for $\alpha-$decays cross the stability region of the nuclei whereas the $p-$ and $n-$ emissions do not.

This was the shortest reasoning I have found. The long version is an entire chapter with detailed calculations, approximations and measurements about the two fundamental forces and their potentials involved.

 

[Dr_Nate]

That's an interesting question. I hope an expert can come by and give a good answer.

In the mean time, from this wikipedia article (https://en.wikipedia.org/wiki/Cluster_decay) the short answer seems to be tunneling. The long answer is likely in the cited works.

 

[Vanadium 50]

These are all categories of spontaneous fission: (N) neutron emission, (N) proton emission, deuteron emission, alpha emission, cluster emission and spontaneous fission. All occur when the energy of the daughters is less than the energy of the parents.

If the question is why 12C emission is rarer than 4He emission, it's because it requires, in effect, three simultaneous alpha emissions all placing their alphas in nearly the same phase space so you get one 12C out. This is rare - and indeed, the question isn't so much why cluster decay is so rare, it's why it happens often enough to observe at all.

 

[Marisa5]

Online I couldn't find much more explanation than "alpha particles have high binding energy per nucleon", so apparently it's quite easy to understand. Any input is appreciated :)

If you take a graph of binding energy as a function of atomic number and flip it upside down, you'll see that helium 4 makes essentially a large potential well at the beginning of the chart. This makes it a state that is more likely to occur than the others because energy has to be added to get out of the well and reach the states further down.

 

[haushofer]

These are all categories of spontaneous fission: (N) neutron emission, (N) proton emission, deuteron emission, alpha emission, cluster emission and spontaneous fission. All occur when the energy of the daughters is less than the energy of the parents.

If the question is why 12C emission is rarer than 4He emission, it's because it requires, in effect, three simultaneous alpha emissions all placing their alphas in nearly the same phase space so you get one 12C out. This is rare - and indeed, the question isn't so much why cluster decay is so rare, it's why it happens often enough to observe at all.

Ok, that sounds plausible, but can I also calculate the probability of this 12C emission as a tunneling phenomenon?

If I find the time I'll put down some math here I did to answer my own question :P

 

[haushofer]

If you take a graph of binding energy as a function of atomic number and flip it upside down, you'll see that helium 4 makes essentially a large potential well at the beginning of the chart. This makes it a state that is more likely to occur than the others because energy has to be added to get out of the well and reach the states further down.

Thanks. The way I now see it, is from the expression of Q; the mass of the helium-nucleus is quite small compared to its constituents, such that Q becomes larger. This energy is mainly used as kinetic energy for the emitted particle.

Sketching a calculation: the tunneling probability (or: the Gamow-factor) depends among others on the mass of the emitted particle, and on the electric charge. Light particles with less charge are more easily emitted. Because the alpha particle is relatively light and tightly bound, the Gamow factor for such an emission is small compared to other emissions, and hence the probability relatively high. I found a similar reasoning here,

http://www.eng.fsu.edu/~dommelen/quantum/style_a/gamow.html#SECTION086114000000000000000

It's more subtle than a lot of textbooks sketch, I suppose.

 

[fresh_42]

It's more subtle than a lot of textbooks sketch, I suppose.

Yes, these are the detailed calculations over Coulomb potentials. An exact calculation involves numerical integrations. And as roughly $G \sim \dfrac{1}{\sqrt{E_\alpha}}$ we're back at comparing energies, and Hanna's (shorter) method.

 

[Vanadium 50]

but can I also calculate the probability of this 12C emission as a tunneling phenomenon?

Sure, but what do you use for inputs?

(Note that you're essentially moving the "close enough in phase space" problem from the final state to the initial state)