# Can an Alpha Particle Fuse with a Hydrogen Atom?

I was wondering if alpha particles created by radioactive decay ever have enough energy to fuse with something else (e.g. hydrogen or another alpha particle).

Staff Emeritus
2021 Award
alpha + p -> 5Li, which is unstable and decays 5Li -> 4He + p. So the protons just bounce off.

mfb
Mentor
They could fuse with a deuteron to Li-6, with C-12 to O-16, and there are many more options, the energy of most decays is sufficient to lead to fusion. All those processes are rare, however, most of the time the particle will lose its energy from interactions with electrons.

ChrisVer
Gold Member
They could fuse with a deuteron to Li-6, with C-12 to O-16, and there are many more options, the energy of most decays is sufficient to lead to fusion. All those processes are rare, however, most of the time the particle will lose its energy from interactions with electrons.

Would that mean that if you could concentrate a lot of alpha-unstable nuclei within a Li-6 container, you could achieve a "Big Bang" [explosion]?
If I recall well, the coulomb potential can be overpassed by particles of several MeV energy.

They could fuse with a deuteron to Li-6, with C-12 to O-16, and there are many more options, the energy of most decays is sufficient to lead to fusion. All those processes are rare, however, most of the time the particle will lose its energy from interactions with electrons.
That's really interesting. If you striped away the electrons could you use alpha particle producers with a short half lives to make a fusion reactor? Maybe having the alpha particles shoot into some kind of ionized gas.

Last edited:
Drakkith
Staff Emeritus
That's really interesting. If you striped away the electrons could you use alpha particle producers with a short half lives to make a fusion reactor? Maybe having the alpha particles shoot into some kind of ionized gas.

No, the cross sections of the reactions are far too small. The vast majority of the time the alpha particle would simply scatter off of the nucleus and not fuse.

No, the cross sections of the reactions are far too small. The vast majority of the time the alpha particle would simply scatter off of the nucleus and not fuse.
Interesting. Thanks for the reply Drakkith. How do you find out the cross section of various possible events, and how exactly is it used to determine probability?

Last edited:
mfb
Mentor
Finding precise numbers can be difficult, but here a rough order of magnitude is sufficient: the fusion cross-section is at most of the order of the size of the nuclei (probably much smaller), while relevant scattering (and therefore energy loss) is orders of magnitude more frequent.

Finding precise numbers can be difficult, but here a rough order of magnitude is sufficient: the fusion cross-section is at most of the order of the size of the nuclei (probably much smaller), while relevant scattering (and therefore energy loss) is orders of magnitude more frequent.
Thanks for the info mfb. Is the cross-section always at most the size of the nuclei?

ChrisVer
Gold Member
Thanks for the info mfb. Is the cross-section always at most the size of the nuclei?

It's a rough estimate for the scattering cross section. The scattering cross section gives the likelihood of a scattering to occur.
Suppose you have some ball of radius $R$ in your sample, and you shoot a ball of radius $r$, how likely will they scatter?
Well, it depends of course on the $2 \pi (R+r)^2$ (the area around a sample ball in which if the incident ball enters, they will "hit" each other). Of course that's a rough /classical estimate.

http://en.wikipedia.org/wiki/Cross_section_(physics)

In general there is no particle physics or atomic physics without cross section, so I'd recommend you to look for it if you are interested in the subject...
In my undergrad I was always told that what you actually want to result in is a cross section...all the bad calculations of QFT have this goal... because that's what is measured.

Last edited:
mfb
Mentor
Thanks for the info mfb. Is the cross-section always at most the size of the nuclei?
For fusion with alpha nuclei, I would be surprised to see a value larger than that.
For nuclear reactions in general, the cross-section can exceed the size of the nucleus by more than a factor of a million. Xenon-135 is like a giant vacuum cleaner for slow neutrons.

Thanks ChrisVer and mfb. No wonder fission is so much easier then fusion.

Astronuc
Staff Emeritus
In the MeV range, scattering is largely potential scattering.

Basically, to measure cross-sections, one would take an ion bean, of alpha particles in this case, and impose the beam on a target material and measure the outcomes. One would do an energy sweep and generate cross-sections derived from scatter and whatever interactions evolved.

In the case of alpha particles, alpha-emitters like Ra, Po, Pu, Am have been mixed with Be to make (α, n) sources. AmBe neutron sources are used for startup sources for reactors, otherwise, spontaneous fission sources, e.g., 252Cf are used.

These sources are sealed, and besides the radiation and radiotoxicity of the radiaonuclides, Be is very toxic in the human body.

More later on fusion reactions.

Last edited:
e.bar.goum
These sources are sealed, and besides the radiation and radiotoxicity of the radiaonuclides, Be is very toxic in the human body.

Absolutely true. Of all the things I handle in the lab, it's not the hot (quite radioactive) sources that give me the pause, it's anything to do with handling beryllium. New gloves, separate tools, respirator, the works. Beryllium gives you Berylliosis, which is quite similar to mesothelioma. This is quite unfortunate, because not only is beryllium quite handy for making a nice neutron source (and means that undergraduates can do neutron absorption measurements), it's a really interesting nucleus for those of us interested in the nuclear physics of light nuclei. It's also rather useful if you want to make beams of light exotic nuclei (6He, 8Li). So I end up coming across it more than I would like.

On the topic of the OP, 5Li has a half-life of about half a zeptosecond, which is on the same order as the timescale of a nuclear collision. I'm now wondering what measurement you would make that would distinguish straight-up rutherford scattering from α+p → 5Li -> α+p. γ-α or γ-p coincident measurements, and/or angular distributions, I suppose. I suspect that you would have your work cut out for you.

Staff Emeritus
2021 Award
5Li has a half-life of about half a zeptosecond, which is on the same order as the timescale of a nuclear collision

Not surprising, since it's unbound. There are no A=5 nuclei.

e.bar.goum
Not surprising, since it's unbound. There are no A=5 nuclei.

I would instead say that there are no stable A=5 nuclei. Deciding what actually constitutes a "nucleus" can be tricky in this mass region. We say that 5Li is a nucleus (insofar as it appears on the chart of the nuclei, and nuclear physicists refer to it as such), but we say that 14Li doesn't exist (beyond the neutron dripline).

You'd properly describe 5Li as a resonance of a+p, but that doesn't stop nuclear physicists describing 5Li as "an isotope" or writing "the 5Li nucleus". Or even writing "5Li has a half-life of about half a zeptosecond, which is on the same order as the timescale of a nuclear collision". Similarly, the 8Be g.s. has a half-life of about 10^-16 of a second. This is obviously very short, but long enough that it exists long past the collision timescale (it's decay is referred to as "asymptotic breakup"). Where do you draw the line as to what a nucleus is? 10^-10 seconds? 10^-3?

ETA: I got curious and checked out the TUNL evaluation for 5Li. And impressive number of identified levels! http://www.tunl.duke.edu/nucldata/figures/05figs/05_02_2002.gif

Last edited:
In the MeV range, scattering is largely potential scattering.

Basically, to measure cross-sections, one would take an ion bean, of alpha particles in this case, and impose the beam on a target material and measure the outcomes. One would do an energy sweep and generate cross-sections derived from scatter and whatever interactions evolved.

In the case of alpha particles, alpha-emitters like Ra, Po, Pu, Am have been mixed with Be to make (α, n) sources. AmBe neutron sources are used for startup sources for reactors, otherwise, spontaneous fission sources, e.g., 252Cf are used.

These sources are sealed, and besides the radiation and radiotoxicity of the radiaonuclides, Be is very toxic in the human body.

More later on fusion reactions.
This is really interesting. Thank you. Could spallation from alpha particle produce neutrons cheaply and in enough numbers to transmute useful amounts of uranium 238 and thorium 232 into fissile isotopes which can then be used as nuclear fuel?

Also, there is something that has been confusing me for a while. I've heard some reference to neutrons in fast reactors fission actinides without being absorbed. Does the fission cross section talk about both actinides which absorb a neutron then fission and actinides which fission without absorbing a neutron?

Absolutely true. Of all the things I handle in the lab, it's not the hot (quite radioactive) sources that give me the pause, it's anything to do with handling beryllium. New gloves, separate tools, respirator, the works. Beryllium gives you Berylliosis, which is quite similar to mesothelioma. This is quite unfortunate, because not only is beryllium quite handy for making a nice neutron source (and means that undergraduates can do neutron absorption measurements), it's a really interesting nucleus for those of us interested in the nuclear physics of light nuclei. It's also rather useful if you want to make beams of light exotic nuclei (6He, 8Li). So I end up coming across it more than I would like.

On the topic of the OP, 5Li has a half-life of about half a zeptosecond, which is on the same order as the timescale of a nuclear collision. I'm now wondering what measurement you would make that would distinguish straight-up rutherford scattering from α+p → 5Li -> α+p. γ-α or γ-p coincident measurements, and/or angular distributions, I suppose. I suspect that you would have your work cut out for you.

Thanks for the info. I was wondering is this possible?

α+H-2 -> Li-6

e.bar.goum
Thanks for the info. I was wondering is this possible?

α+H-2 -> Li-6

Sure. It's possible. Why wouldn't it be?

That's actually the major reaction that goes towards producing 6Li in the big bang. (Not that 6Li is produced in any serious abundance, but that's the reaction that produces it).

Sure. It's possible. Why wouldn't it be?

That's actually the major reaction that goes towards producing 6Li in the big bang. (Not that 6Li is produced in any serious abundance, but that's the reaction that produces it).
Thanks for the info. I don't know much of the theories involved so I'm not to clear on what is or isn't possible.

e.bar.goum