10
$\begingroup$

Is there a known reason why any given element has finitely many isotopes? Here I mean both stable and unstable isotopes.

If we know this, do we have a reason why, for a given element, are the isotopes limited to that particular number?

$\endgroup$
15
  • 3
    $\begingroup$ An infinite number of isotopes would require that some of them have extremely large (like macroscopic) mass. To get "infinitely many, you'd need some that had arbitrarily large numbers of neutrons. Surely that cannot be what you meant to ask with this question? $\endgroup$ – Brick May 28 at 14:20
  • $\begingroup$ well, I dont mean actual infinity, I mean what is the upper limit and why is there one. Also for an element with n protons do we always get isotopes of the form n, n+1, ..., n+m for some m? $\endgroup$ – Frank Conry May 28 at 14:48
  • 2
    $\begingroup$ Are you familiar with this? $\endgroup$ – J.G. May 28 at 14:54
  • 2
    $\begingroup$ Search term: neutron/proton drip line. $\endgroup$ – rob May 28 at 15:05
  • 2
    $\begingroup$ I'm definitely no physicist, but I think that if you manage to put enough neutrons together, you end up with a black hole. That would give you a hard (finite) limit. $\endgroup$ – Martin Argerami May 29 at 4:46
13
$\begingroup$

I think this is a good question -- after all, if there's no extra Coulomb repulsion penalty for adding more neutrons, unlike for protons, why can't nuclei have lots of neutrons?

One model for the nucleus we use is called the Semi-Empirical Mass Formula (SEMF), which has a bunch of terms describing the energy contributions to the nucleus. See wikipedia for the full formula. The main term that answers your question is the "Asymmetry Term", given by $$a_\text{A}\frac{(N-Z)^2}{A}$$ where $a_\text{A}$ is some constant we can find empirically, $N$ is the number of neutrons, $Z$ is the number of protons and $A=N+Z$ is the nucleon number.

This is a penalty term in the energy of a nucleus. If there is a large difference in $N$ and $Z$, this term is large. If $N$ is similar to $Z$, the term is not as large. The rational for this is the Pauli Exclusion Principle, which tells us identical particles cannot occupy the same energy state. If we're adding lots of identical neutrons, we must put them in different energy states. We can get a cheaper energy cost by filling in some protons instead for a given nucleon number $A$.

To answer your question in the comments: why do isotopes often have more neutrons than protons, I think the answer there is it is somewhat favourable to add nucleons to the nucleus, because that increases the strong force present, but its cheaper to use neutrons than protons, at least for cases where the ratio $N/Z$ does not deviate too far from 1.

$\endgroup$
2
  • $\begingroup$ I assume there is a experimental correlation whereby half life decreases as N/Z increases but is there any threshold there where N/Z is so high the nucleus cannot form? $\endgroup$ – Frank Conry May 28 at 16:57
  • 2
    $\begingroup$ I don’t think it’s a question of whether a nucleus can or cannot form, more a question of how many energetically more favourable alternatives there are for the bunch of protons and neutrons to form. As mentioned in another answer, there’s no reason 9H cannot exist, but its energy for the given value of A would be high and it would want to decay very quickly. $\endgroup$ – Garf May 28 at 17:01
7
$\begingroup$

In the comments you clarify your question by asking, as an example, about hydrogen isotopes.

If you look at a Table of Nuclides, you will see that there are at least 7 hydrogen isotopes which have been identified so far. There are links attached to each entry in the table that give data on the reactions for creating these exotic nuclides.

You can see that the He-3 through He-10 have been identified, C-8 through C-22. At the extremely neutron-rich ends, the halflives are extremely short, and neutron emission is prevalent.

It seems, based on the experimental data, that the only restriction on neutron-rich isotopes of any element are the experimental ability to make the nucleus long enough/in sufficient number to get repeatable measurements to demonstrate they have actually been made.

EDIT: As pointed out in a comment, there most likely is a limiting halflife. Check out this question and answer.

There's no reason to assume that H-8 and H-9 won't eventually be formed and identified. It will take a lot of ingenuity, patience, and money.

$\endgroup$
6
  • 2
    $\begingroup$ There has to be a lower limit to the half-life, though. As mentioned in physics.stackexchange.com/a/468493/123208 "the theoretical limit on very short half-lives is the fact that information cannot travel faster than the speed of light. Obviously, a nucleus must be bound before we can discuss its half-life. But it makes no sense to say that a nucleus is bound if it decays before all of its constituent protons and neutrons "know" that it has formed". $\endgroup$ – PM 2Ring May 28 at 15:54
  • $\begingroup$ @PM2Ring I see the point made there so I guess there is some type of bound. But there is no nuclear model predictor for when that limit is reached. Would it be H-13, H-20, ? I expect it might be before those, but I don't know of a model to calculate what adding another neutron will do. $\endgroup$ – Bill N May 28 at 16:11
  • $\begingroup$ @PM2Ring It's also interesting to me that the chart I linked (presumably, the latest information available) doesn't have He-2, which some consider a real intermediate step in the proton-proton cycle, just before it decays to H-2. I guess no one on Earth has measurably identified He-2. $\endgroup$ – Bill N May 28 at 16:13
  • $\begingroup$ @PM2Ring Answer has been edited. Thanks for the link. $\endgroup$ – Bill N May 28 at 16:16
  • $\begingroup$ Yes, it's hard to measure the half-life of He-2 (aka diprotium). It's hard to form it, and even when you do manage to make it, it almost immediately falls apart, which doesn't give the slow weak force much time to convert it to deuterium. As I recently said here, the probability of that conversion is of the order of $10^{-26}$. $\endgroup$ – PM 2Ring May 28 at 16:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.