Even with that short half life, it’s still a bound state. If you try to break it into individual neutrons and protons, you will have to spend energy. This isn’t true for any collection of protons and neutrons.
Consider a flat surface with a concave dent in the middle. If you roll a marble towards the middle, the marble can get stuck in the dent – in a bound state – if the system can get rid of some of the energy gained from the fall into the dent. (In a nuclear system, this might be gamma ray emission. In a chemical system, this would be generic “heat” energy or eV-scale photon emission.) In any case, there is a bound state available.
Now say the middle of the surface is actually flat or has a small bump. You won’t be able to get the marble to stop in the middle. There is no bound state there.
Now imagine in the first case that the marble is resting at the bottom of the dent. Imagine that a few inches over, there is an even deeper dent. If the marble could slide over to the other dent, it would, as the other dent is a more energetically favorable place to be. But, alas, the marble is stuck. In a quantum mechanical system, however, the “marble” can move to the other dent via tunnelling. The ability of a system to tunnel from one state to another without the input of energy is why bound states can be unstable. How hard the tunnelling is determines the lifetime of the bound state.
Many proposed grand unified theories allow or even require proton decay, but within the standard model proton decay is forbidden.
As for long-lived vs. stable nuclides, there’s the question of what something can decay into. If the total energy (including mass-energy) of all possible daughter systems is higher than the mass of the nuclide in question, it is stable. This assumes we are considering all possible daughter systems, but that is a question of whether there are decay processes missing from the theory rather than a question on the definition of “stable”.
Particles and nuclei for which decays are allowed but for which no decay has been observed would not be called “stable” but would instead have a lower limit quoted for their lifetime.
My understanding in high school chemistry & physics was that EVERY isotope for any element will eventually decay, if you’re willing to wait around for trillions and trillions of years. Is this true?
Only if proton decay is possible (which we don’t know), and even then only after much longer than trillions of years. (The proton lifetime is no smaller than 10[sup]35[/sup] years. That’s waiting a trillion years a trillion times in a row and then doing it all again another trillion times (almost).)
That’s something I look forward to finding out. Hell, some protons are already 13 billion years old, so what’s another 9.99999999999999999x10[sup]34[/sup] years?
Sheesh, you don’t just wait 10[sup]35[/sup] years staring at one proton. You get 10[sup]35[/sup] protons and wait one year. Let’s see, 1.6×10[sup]-27[/sup] kg per proton so you need less than 10[sup]8[/sup] kilograms of protons. Pure H-1 would do. (Don’t want to fiddle with those messy neutrons.) And isolate form all other possible source of radiation. Let’s get cracking Dopers. Fame and a Nobel prize awaits!
This is an overstatement. Life has evolved in the presence of water fored predominantly from [sup]1[/sup]H. There’s no reason to think that life isn’t possible in water made from all [sup]2[/sup]D. Any such life would have a similar problem surviving if it were suddenly placed in an [sup]1[/sup]H-only environment.
A bit of an aside, but there have been people looking at whether Tetraneutrons, a bound state of four neutrons and no proton, exist. From the link,
This seems odd to me, though, since I’d expect the strong nuclear force to make the tetraneutron be at least a little more stable than four isolated neutrons (which should have a half-life of about 4 minutes, decaying to three neutrons and a proton).
A good many theories have been ruled out, but a good many are still allowed. Besides, there’s always wiggle room. Never underestimate the creativity of theorists in evading experiemental bounds.
The nuclear force is a bit more complicated than just “attact”. In particular, it is strongly spin-dependent, with the sign such that spin singlets gain a repulsive term and spin triplets gain an attractive term. Considering the two-nucleon case: n-n (or p-p) differs from n-p (deuteron) in that the Pauli exclusion principle is relevant for the former. If the neutrons’ spins are aligned, you lose big time due to the necessity of having antisymmetric spatial wavefunctions. (You get a “hard” core repulsion.) If the neutrons’ spins are anti-aligned, you evade the exclusion principle, but you lose through the spin-dependence of the nuclear force. In the end, n-n is not bound at all. (By extension, neither is the spin singlet n-p state. The n and p spins are always aligned in a deuteron.)
An as-yet unreproduced experimental result has suggested that tetraneutrons exists, motivating nuclear theorists to try to crowbar in modifications that would yield a bound four-neutron state. Current understanding and many corroborating experimental results, however, strongly suggest the nonexistence of tetraneutrons.
Well, yes. Once you get the black hole down to a really low mass, you can get pretty much anything out of it. But you’d need to emit a heck of a lot of lightweight stuff before you got down to that point.
Here’s a question about the table of nuclides. Why is it organized so that isotopes of the same weight are on a diagonal? S34 is to the left and above P34 which is to the left and above Si34. This is opposite the trend of the chart which rises from the left to the right. Wouldn’t it be more sensible to organize it so that all the isotopes of the same weight are above each other? S34 directly above P34 which is directly above Si34, etc… It seems like it would be easier to find the item you want since that way each element is on its own row and each weight is on it’s own column. And the chart would still have the rising towards the right trend, if steeper.
It’s so you can follow radioactive decay chains. The skewing makes it easier to follow alpha decay, and other forms of decay move you to an adjacent cell or on a diagonal.
