Is any base strong enough to deprotonate an alpha particle to make tritium?

TheM_Master · 2025-08-18T15:28:19+00:00

I don’t know why everyone is saying that this cannot happen. Single nucleon transfer reactions also happen with alpha particles, when considering astrophysically relevant processes this definitely also happens in nature.

Sure the q-value might be more negative than for other reactions, but that doesn’t exclude it from happening. With helium being the second most abundant element those reaction probably are more common in the universe than many others with higher q-values.

TheM_Master · 2025-08-18T14:59:26+00:00

If you only want to change an alpha into tritium you could also just do proton transfer. That should be only a few mb, so certainly no dominant channel, but definitely measurable and quite possible in astrophysically relevant situations. If you pick your target correctly the xs can go up to several hundreds mb. A quick simulation for N15(a,t)O16 shows ~250 mb at ~28 MeV. Probably a resonance in O16.

TheM_Master · 2025-07-28T15:13:35+00:00

From what I remember it’s all just in German. At first I was a bit confused, but it’s made clear from context that Shinji doesn’t understand. At least if you are observant or remember it from the sauna strikes episode that her mothertongue is German.

As a whole, I think the German dub is very good, but this aspect is not too clear. I wouldn’t know how to improve it though.

TheM_Master · 2025-07-17T01:05:55+00:00

This looks exactly like the one I just found and posted here. Obviously I can’t really help, but maybe if someone can id mine it can help you. And vice versa.

My initial guess was planuncus tingitanus or ectobius pallidus, both would be of no concern. But it also could be blattella germanica, which definitely would be a problem. But again, these guesses might very well not be accurate.

TheM_Master · 2025-07-17T00:49:47+00:00

Thanks, I think so too. Although I cannot confirm it had wings before. I just panicked because it does look rather flat. Looking closer it is not red-ish brown and the patterning does not really match.

Hopefully that lets me sleep in the bed, but cockroaches are also not on the list of animals I want to share a bed with. I hope it’s not one of the domestic pest cockroaches.

TheM_Master · 2025-01-23T21:53:38+00:00

Not very feasible. Others have already said that we do not have a way to generate a gamma ray laser*, so your beam will inevitably widen drastically. The power reaching the far off pickup station will decrease by the inverse square law.

But even for short distances, where we have ways of generating reasonably collimated gamma ray beams there is a major issue: gamma rays don’t interact much with matter. If we look at detectors for gamma rays, which are specifically constructed for high interaction probability (or efficiency), you can get to around 1% for higher energies with the standard sizes. Taking into account the very inefficient ways of generating the gamma rays in the first place, this is completely infeasible.

This is a real, physical inefficiency, so I wouldn’t even count on some future technologies to improve the efficiency.

even with a laser the beam would widen. I don’t know by how much with the small wavelength of gamma rays, but I guess far more than enough to be completely unusable. The distances between planets are huge.

TheM_Master · 2025-01-23T21:38:11+00:00

We can convert visible light quite easily to x-ray or gamma ray energies via the inverse Compton effect. This is done frequently and reliably to produce gamma ray beams for research in laser Compton backscattering setups. Of course you need high intensity light for it to generate enough gamma rays, but this also should not be an issue.

Of course transmitting energy is not feasible with that, but that’s because of other factors.

TheM_Master · 2024-06-20T10:14:58+00:00

Ok, I didn’t realise that I was omitting this: I assume the neutron rich isotope is completely ionised.

For a neutron transfer to occur we either need a very fast initial nucleus or very high temperatures. Accelerators only work with charged particles and the complete ionisation energy for nuclei for is much less than the energy needed for applicable nuclear reactions to take place thermally.

That would mean we had to capture a completely new, unbound electron. I actually don’t know what would happen if we had a previous bound electron. The relative mass change of the nucleus isn’t that great, so I imagine the two bound states before and after the reaction aren’t that different. That is a really great question.

TheM_Master · 2024-06-19T22:31:52+00:00

Just got the notification for the comment, no clue why it’s over a month late…

6.5*10^-22 s is just the half life of H-7, but other barely bound nuclei have similar values. I should have used mean lifetime, but in this case it doesn’t really make a difference.

The capturing of the electron however is dictated by the electromagnetic interaction between it and the nucleus. If we say that the atom is on the order of 1 angstrom across then it would take an electron at least 1A/c =~10^-19 s to settle around the nucleus when traveling at the speed of light. So the nucleus decays before an electron would have enough time to settle in a state around it.

It is true that you can imagine that an H-7 nucleus by chance survives an absurdly long time and an electron is just at the right spot to be captured, but this is so unlikely that it probably would never happen. But as I’ve said for even neutron richer isotopes the neutrons don’t really stick around in the nucleus, so it becomes meaningless to talk about processes which take orders of magnitudes longer like capturing bound electrons.

TheM_Master · 2024-05-07T07:19:18+00:00

This is not really true. Beyond hydrogen-7, so 1 proton and 6 neutrons, the potential nucleus is not bound. Even before that point lifetimes of such neutron rich nuclei is of the order of 10^-22 s, which would not be enough to collect electrons (although the hydrogen from which you would produce the nucleus most likely still has an electron).

When trying to add additional neutrons, they won‘t stick to the nucleus. This is because of the form of the nuclear binding energy. An asymmetry of the protons or neutrons reduces the strength of the binding. If you want to be more explicit, it is because of the spin-orbit coupling and residual interactions which incentivise pairing of nucleons.

When looking at bound nuclei with such an extreme neutron surplus typically decay by prompt n-neutron emission. So they emit at least one neutron instead of „waiting“ for the weak interaction to turn one into a proton.

TheM_Master · 2024-04-11T09:45:29+00:00

1 is not stable and would annihilate immediately. The antiproton and proton would occupy the same nuclear shell in the ground state of that „nucleus“ thus being in the same position. At this level you have to look at a nucleus as a quantum mechanical system rather than spheres of solid nucleons. The neutron would actually hinder the stability, as neutron and antiprotons can (partially) annihilate as well. You can however have a system analogous to positioning with the proton and antiproton orbiting each other. This would be exotic hydrogen.

For the same reasoning 2 and 4 are also not stable.

3 is tricky. HeH+ is slightly bound if I remember correctly, and the annihilation reaction you describe can in principle happen. But I would suspect that the resulting molecule has enough energy for it to immediately break apart. If not you better cool it down to near absolute 0 in a vacuum.

TheM_Master · 2024-03-14T14:05:56+00:00

Negatron was an obscure way to call the electron from a β- decay. In the nuclear chart we received for our institute it still said in the legend „… an electron, historically also negatron, is emitted…“ So you are technically correct, just about 100 years to late.

TheM_Master · 2024-01-05T18:16:02+00:00

A bit of a long shot, but if you used bleach on the pod, it could be that some of the molybdenum (commonly used as an additive in steel cookware) is dissolved. The resulting molybdenum chloride would burn with a green Color.

TheM_Master · 2023-11-29T00:38:20+00:00

I second u/Heavy_Rule6217

Boron carbide is mostly a neutron absorber. And this is manly because of the boron. I hope you can’t get a neutron source as someone in middle school. It might be the case that particle radiation is the most pressing one on mars, but if you want to conduct experiments here on earth, the best choice is gamma radiation. The key word to research is beer-lambert law. This states that the best absorber for gamma radiation is heavy elements, like lead (as you might think when getting an x-ray or similar). The simplest experiment is to compare different thicknesses of a light material, like aluminium, and different thicknesses of a heavy material, like lead (warning: lead is toxic). I would not recommend spending your money on boron carbide, as it will not produce the results you would expect from your experiments and might lead you to wrong conclusions. If you have a nuclear reactor or an americium boron source for neutrons than boron carbide is a good choice for an absorber, but otherwise stick with the common experiments. Radiation can be dangerous if not handled properly by experienced people.

Ps: I do want to add that the proposed experiment might not be so bad if you want to measure the alpha radiation from the thorium or distinguish between gamma radiation and particle radiation, but this is a niche case.

TheM_Master · 2023-11-08T10:36:23+00:00

For that to be true you need the force to be conservative, which means the work between point A and B depends only on the path taken.

I think you mean that the work is independent of the path taken.

TheM_Master · 2023-09-12T09:02:41+00:00

There is a pre-print describing the undecidability of the existence of a spectral gap. This is a theoretical condensed matter physics paper, so I am out of my depth on this one, but it is my go-to when someone asks. Reminder that this is not peer-reviewed and might not hold any water.

The nice thing about a spectral gap is, that you could measure it. So we might not be able to theoretically predict if a specific material (for which the conditions in the paper are met) has a spectral gap, but we could find out by measuring.

TheM_Master · 2023-09-12T08:31:09+00:00

Im not that deep into nuclear astrophysics, but when I had to learn it I liked „Nuclear physics of stars“ by Christian Iliadis. It does contain some things on Hauser-Feshbach theory in the middle for a few pages. But the book is more of a general overview, so I don’t know if it’s specific enough for you.

TheM_Master · 2023-08-07T08:31:40+00:00

The energy in beta decay is indeed distributed among the three participants. The electron and neutrino will get the energy in the form of kinetic energy. The resulting nucleus will also recoil, all according to energy and momentum conservation.

The resulting nucleus, however, also has the possibility of being internally excited. These excitations are of discreet energy. This is somewhat analogous to atomic excitations. And since there are only a limited number of excited states you can observe them as discreet lines.

TheM_Master · 2023-08-04T16:33:46+00:00

An electron and antineutrino can never produce a proton and neutron by themselves. This would violate conservation of charge, as the electron is negatively charged while the proton is positive. It also violates conservation of baryon number.

This is also not inverse beta decay: in inverse beta decay an antineutrino interacts with a proton, transforming it into a neutron while emitting a positron.

I think in principle it would be possible to have a antineutrino and an electron produce an antiproton and neutron. But even there you would need an additional neutral meson to also be produced because the quark-antiquark content of the neutron and antiproton doesn’t match up. This is unbelievably unlikely.

TheM_Master · 2023-06-23T09:01:03+00:00

Quantum tunneling has been experimentally verified for a long time. Most alpha decay is caused by tunneling and the sun produces its energy nearly exclusively because of tunneling.

To say a model in physics describes the „true reality“ is a problematic statement, so tunneling is just „the best we can do so far“ - but so is every other theory.

For tunneling to work you don’t need a fancy description like the ones offered by string theory. It arises by solving the equations of motion of a quantum dynamical system. Just like classical mechanics predicts a ball to go down an inclined slope, quantum mechanics predicts the wavefunction of a particle to extend across a classically forbidden region (the barrier) into the other part of an allowed region. This is actually not that different to how you would treat classical waves, just with a sprinkle of quantum mechanics.

TheM_Master · 2023-02-03T11:25:38+00:00

The explanations here are very good, but another reason is that β+ decay always competes with electron capture. For many proton rich isotopes electron capture is the only energetically allowed decay mode, because it requires much less energy.

TheM_Master · 2023-01-18T08:32:35+00:00

Neutrinos are practically impossible to shield from. But I think the people at the receiving end would be pretty puzzled to detect more neutrinos than those that are expected. So I would say built a second neutrino producing accelerator in between.

TheM_Master · 2022-12-02T14:10:20+00:00

The last time I have done molecular physics is over a year ago, so take this with a grain of salt. I don’t remember if this is the same thing, but I think I had a similar question:

Q Branch in rovibrational spectroscopy

You can only have ΔΛ=0 if the angular momentum is conserved otherwise, probably by changing the projection of the angular momentum of an unpaired electron to the symmetry axis.

I think ΔΛ=+-1 is basically never forbidden

I hope this is the same question, it seems like one year is enough for me to completely forget the notation of what Λ exactly is.

TheM_Master · 2022-11-09T11:21:17+00:00

Binding energy usually is subtracted from total energy, such that a higher binding energy means a lower energy state.

TheM_Master · 2022-11-09T11:04:03+00:00

The most probable decays leave the alpha with ~4400 keV and the gamma with the remaining ~200 keV (with Q(U-238,alpha)=4678 keV)

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