# Identity of alpha particles

1. Nov 25, 2013

### Choisai

As a physics student, I understand that α particles are emitted and are the same as helium atoms without electrons.

But this raises two questions to me:

1) What happens with the electrons at the atom that emits them? He now has two electrons too much. What happens with those?

2) Why an helium atom? Why not nitrogen, hydrogen, barium or whatever other element? All teachers always told me 'it emites helium atoms without electrons. These are known was alpha particles'. But why helium? What equation shows that the particles must be helium?

2. Nov 25, 2013

### mathman

1) The electrons get lost in the mix. Eventually the alpha particle needs to pick them up to become neutral.
2) Nuclear physics theory makes it so.

3. Nov 25, 2013

### Bill_K

Alpha particles are especially easy to form in nuclear reactions, for two reasons: a) they're small (just four nucleons) and b) they're tightly bound (28 MeV binding energy).

4. Nov 26, 2013

### Choisai

"Alpha particles are especially easy to form in nuclear reactions, for two reasons: a) they're small (just four nucleons) and b) they're tightly bound (28 MeV binding energy)."

That is true, but there are many other things that are small too. Hydrogen for an example is small. And the idea that two protons and two neutrons can be put together because they are so tightly bound makes me wonder. Why do they have to be bound together? What equation shows this? My textbook doesn't go any deeper than 'it emits alpha particles, which are helium atoms without electrons' and then goes on how you can calculate with that. Why does it need to emit a single particle that is tightly bound and has to be small?

Since I don't have any equation or mathematical description of how it really is, I am wondering why not heavier elements are produced. Why not a boron-atom without electrons for an example?
Helium is light sure, but why does it need to be light?

5. Nov 26, 2013

### Bill_K

(Forgot to mention, c) Alpha particles have an even number of protons and neutrons.)

You'll find nuclear physics different from other areas of physics, Choisai. The nucleus is a rather complicated object. Fundamentally it is described by the same laws of quantum mechanics, but applying those laws involves large numerical calculations. Simple understanding requires approximate models, and in many cases qualitative arguments. You won't find an "equation" for everything!

In any decay or reaction there's a certain amount of energy available, which must be split into the energy required to create the outgoing particles and their kinetic energy. The more energy it takes to create them, the less there is available to go into their kinetic energy. The less kinetic energy they have, the smaller the phase space. And the smaller the phase space, the less likely the decay. So in general, decays which produce the lightest particles are favored. That's one reason why the tightly bound alpha particle is often emitted.

Other factors have to do with the binding energy of the nucleus itself. Nuclei with even numbers of protons and neutrons have more binding energy. Also the ratio between the number of protons and neutrons is important. It might look like beta decay would always be preferred over alpha decay, but beta decay involves changes from odd to even, or vice versa, which costs energy. Alpha decay therefore mostly happens in cases where beta decay cannot.

6. Nov 26, 2013

### Choisai

That sounds reasonable. Do you have the link for a paper to back that information up so I can read some more on the subject?

7. Nov 26, 2013

### Staff: Mentor

This sort of material is often covered at an introductory level in "introductory modern physics" courses and textbooks, for second or third year university level in the US. Try a Google search for "semi-empirical binding energy formula." Descriptions of it will probably touch on at least some of the things that Bill mentioned, and you may find some discussion specifically about the alpha particle, which is something of a "special case".

8. Nov 26, 2013

### Staff: Mentor

Proton (=hydrogen nucleus) emission is a rare, but possible decay process. The emission of other nuclei (like carbon) is even more exotic, but some nuclei can do that: lists of experimental results

9. Nov 26, 2013

### bcrowell

Staff Emeritus
Helium has 2 protons and 2 neutrons. A particle number of 2 is especially stable in a nucleus. This is because of the shell structure, and it's analogous to the stability of the noble gases in atomic physics. For more on this, google "magic number."

10. Nov 27, 2013

### snorkack

And merely saying "helium" is grossly misleading. Alpha particle is specifically helium 4.
Helium 4, like helium atom, has extreme stability because of closed shells. 2 electrons of opposite spins both in their lowest state, 2 protons of opposite spin ditto, 2 neutrons of opposite spin ditto.
Nothing else matches that stability. Regarding electrons, the other noble gases, neon et cetera, do have noble behaviour, but still they have slightly lower ionization energy than He. Regarding nuclei, there are the higher magic numbers - but still they are not as stable as alpha particle.

As for lack of proton decay, I think it can be explained by lack of a good barrier. If a nucleus is unbound with respect to proton removal then the proton tends to leave really, really fast. Like "lithium 5". The unwanted proton that does not fit into the shells of the alpha particle will bounce off in the region of 10^-22 seconds, so no nucleus to decay in the first place. And the nuclei with surplus protons that do not immediately bounce off tend to have some place for a neutron, so they undergo beta plus decay or electron capture, like boron 8 or beryllium 7.

11. Nov 28, 2013

### bcrowell

Staff Emeritus
No, if that were the reason then when we varied the Z of the emitted particle, we would get a monotonic trend. We don't: Z=1 and Z=3 are much less likely to be emitted than Z=2. (The WKB tunneling probability also depends exponentially on the square root of the mass, but again, this is a factor that would cause a monotonic trend.) The reason that alpha decay is so much more common than emission of Z=1 or Z=3 is because the strong binding of the alpha gives a high Q value for the reaction, which gives much more phase space for the final state.

Your explanation would be more valid as an explanation of why we don't see neutron emission in natural radioactivity from cold nuclei.

Last edited: Nov 28, 2013
12. Nov 29, 2013

### snorkack

Note that small nuclei tend to have problems with alpha decay as well, again thanks to low barrier. Beryllium 8 has alpha decay energy of mere 188 keV, yet decays as fast as 10^-16 seconds.

If the magic number of alpha caused the preference for alpha decay, there should be decays to the other magic number nuclei. Carbon 12 (closed subshell) and oxygen 16 (fully magic). There are not. Radium 226 and 224 do, for some reason, emit carbon 14 (doubly magic yet unstable), and thorium 228 emits oxygen 20, also unstable and a nonmagic neutron number.

13. Nov 29, 2013

### Staff: Mentor

I don't see that logic. There could be a preference for magic numbers, but those decays don't have to exist just because the numbers are magic.

In addition, there might be a selection effect in the measurements: decays to radioactive isotopes could be easier to detect, as there is less natural background.

14. Nov 29, 2013

### Astronuc

Staff Emeritus
8Be is unstable.

Light nuclei may undergo (n, α) or (n, p) reactions, and there are possibilities of (n, 2α) reactions, but they require high neutron energies.

For radioactive decay, one does not see spontaneous alpha emissions for elements lighter than Bi.