Why does the nucleus become unstable as the number of nucleons increases?

[PainterGuy]

Hi,

I'd appreciate it if you could help me with the queries below. Please try to keep it as simple as possible because I have only basic understanding.

What is a radioisotope?

On an earlier page covering isotopes it was learned that isotopes are variants of an element that, while all having the same number of protons, have differing numbers of neutrons. These variants are called isotopes. Because the like charges of the protons repel each other,there are always forces trying to push the atom nucleus apart. The nucleus is held together by something called the binding energy.

In most cases, elements like to have an equal number of protons and neutrons because this makes them the most stable. Stable atoms have a binding energy that is strong enough to hold the protons and neutrons together. Even if an atom has an additional neutron or two it may remain stable. However, an additional neutron or two may upset the binding energy and cause the atom to become unstable. In an unstable atom, the nucleus changes by giving off a neutron to get back to a balanced state. As the unstable nucleus changes, it gives off radiation and is said to be radioactive. Radioactive isotopes are often called radioisotopes.

All elements with atomic numbers greater than 83 are radioisotopes meaning that these elements have unstable nuclei and are radioactive. Elements with atomic numbers of 83 and less, have isotopes (stable nucleus) and most have at least one radioisotope (unstable nucleus). As a radioisotope tries to stabilize, it may transform into a new element in a process called transmutation. We will talk about transmutation in more detail a little later.

Source: https://www.nde-ed.org/EducationResources/HighSchool/Radiography/radioactivity.htm

The strong nuclear force is created between nucleons by the exchange of particles called mesons. This exchange can be likened to constantly hitting a ping-pong ball or a tennis ball back and forth between two people.

Source: https://aether.lbl.gov/elements/stellar/strong/strong.html

Question 1:
Strong nuclear force comes from nucleons therefore more number of protons and neutrons (i.e. nucleons) means more stronger "strong nuclear force". There is only one repulsive force in a nucleus and that is repulsive force due to protons but that repulsive force is very much weaker compared to strong nuclear force. As it is said above that as the number of nucleons go over 83, the nucleus starts becoming unstable. Why is it so?

Question 2:
It also says "In an unstable atom, the nucleus changes by giving off a neutron to get back to a balanced state." Is the statement always true? I don't think an unstable atom always changes by giving off a neutron. For example, 226- radium with atomic number 88 decays into radon-222 with atomic number 86 with emission of an alpha particle, i.e. two protons and two neutrons.

Helpful link:
Periodic table with number of neutrons: https://hobart.k12.in.us/ksms/PeriodicTable/energy levels.htm

 

[Vanadium 50]

There is only one repulsive force in a nucleus

That is not a question, it is a statement. And it is wrong. The strong nuclear force is attractive in some cases and repulsive in others.

 

[PainterGuy]

That is not a question, it is a statement. And it is wrong. The strong nuclear force is attractive in some cases and repulsive in others.

Thank you for pointing this out. Then, could it be said that as the number of nucleons increases, the force increases and then starts becoming repulsive, and this in turn destabilizes the nucleus?

Could you please comment on Question 2?

Thanks for your time!

Helpful link:
https://qr.ae/pNYwjH

 

[Vanadium 50]

I don't want to get involved in a proxy war with another site, sorry.

 

[PainterGuy]

I don't want to get involved in a proxy war with another site, sorry.

I'm sorry but if trusted that information, I wouldn't have asked you for help. I shouldn't have mentioned that link as "helpful"; that was a mistake.

 

[Janus]

Thank you for pointing this out. Then, could it be said that as the number of nucleons increases, the force increases and then starts becoming repulsive, and this in turn destabilizes the nucleus?

No. It would probably be better to think of it like this: The strong force fades off very rapidly with distance, much faster than the electromagnetic repulsion does. This means only nuclei near each other have a strong "grip" on each other. As the nucleus grows, nucleons have a harder time holding on to "further" nucleons against the growing electromagnetic repulsion. But this is only one part of the whole picture.

Could you please comment on Question 2?

Unless is is undergoing fission, a nucleus during radioactive decay doesn't emit a neutron. Instead, a neutron in the nucleus decays into a proton and electron, with the electron being ejected as a beta particle. Free neutrons are not stable, and will decay with a half-life of ~15 min. In the presence of protons in a nucleus, they can be stable. While the presence of neutrons in the nucleus are needed to hold the protons together, the protons also lend stability to the neutrons.
Certain proton/neutron ratios are more stable. Too many neutrons tends to cause beta decay (which is why you just can't add more and more neutrons to a nucleus without limit in order to "thin out" the protons' mutual repulsion), while too few can cause alpha decay ( the ejection of a helium nuclei) There is another option, "Beta+" decay, this is when a proton in the nucleus emits a positron and becomes a neutron in order to restore "balance". [/quote]Thanks for your time!

Helpful link:
https://qr.ae/pNYwjH
[/QUOTE]

 

[PainterGuy]

Unless is is undergoing fission, a nucleus during radioactive decay doesn't emit a neutron. Instead, a neutron in the nucleus decays into a proton and electron, with the electron being ejected as a beta particle. Free neutrons are not stable, and will decay with a half-life of ~15 min. In the presence of protons in a nucleus, they can be stable. While the presence of neutrons in the nucleus are needed to hold the protons together, the protons also lend stability to the neutrons.
Certain proton/neutron ratios are more stable. Too many neutrons tends to cause beta decay (which is why you just can't add more and more neutrons to a nucleus without limit in order to "thin out" the protons' mutual repulsion), while too few can cause alpha decay ( the ejection of a helium nuclei) There is another option, "Beta+" decay, this is when a proton in the nucleus emits a positron and becomes a neutron in order to restore "balance".

Thank you very much for the simple reply.

I think in the discussion about nuclear reactions one should be careful with the words "giving off", "transforms", "emits", "ejects", etc. I think in the original statement about Question 2 the word "transforms" would have been a better choice.

Relevant links:
/watch?v=9b8qZ6OHZ5s (put youtube.com in front)
/watch?v=gqrh8wbPXVE
/watch?v=3koOwozY4oc
/watch?v=fES21E0qebw

 

[snorkack]

Counting "forces", remember that besides the repulsive part of strong force and electromagnetic force, there are also repulsive "forces" of zero point oscillation and exchange repulsion.
Generally nuclei hardly ever stabilize themselves by giving off neutrons - or even protons. The reason is that when the nucleus is on neutron or proton dripline, the neutron is bounced off immediately, on elastic collision timescale. Giving off a neutron happens either at once (and then isotope does not really exist) or not at all.

Nuclei that contain too many neutrons or protons generally stabilize themselves by turning neutrons into protons, or vice versa. Neutrons can turn into protons by emitting electrons (beta decay) while protons can turn into neutrons by emitting positrons or capturing electrons.

An important term here is "isobar". Isobars are nuclei that have same number of protons and neutrons together, but different numbers of each. Isobars are by definition different elements (because different number of protons).

An easy rule to note is Mattauch isobar rule. It has two notorious violators (Te-123 and Ta-180) but still usefully illustrative.
Mattauch rule says that if two isobars differ by one proton, at least one of them is radioactive.
The science advisor Vanadium 50 is himself almost a third Mattauch violator with half-life of 1,4*10¹⁷ years.

 

[anorlunda]

This means only nuclei near each other have a strong "grip" on each other.

Did you mean "This means only nucleons near each other have a strong "grip" on each other."?

 

[mgeorge001]

I think it likely that this type of question can benefit from some historical perspective. The modern physics involved with nuclei did not develop in just one step, and, in fact, benefitted from insights that nuclear physicists gained from theories and experiments in other fields. You may wish to ignore my post as I am merely going to give some references to the very early work, which is extremely revealing, in "classical" quantum mechanics, before the rise of sophisticated qed and qft after Feynman, et al. First, I recommend the following Wikipedia articles: Island of stability; Nuclear physics; and Proca action. Second, there is a really nice historical overview about the significance of Proca's work: "Proca equations of a massive vector boson field", by D.N. Poenaru, who, as oriented toward Romania, emphasizes Proca's achievement very strongly and puts it into a really good perspective! This will only help a little: There have been numerous advances in nuclear physics since Proca's day. Thanks for the great question.

 

[snorkack]

In most cases, elements like to have an equal number of protons and neutrons because this makes them the most stable. Stable atoms have a binding energy that is strong enough to hold the protons and neutrons together. Even if an atom has an additional neutron or two it may remain stable. However, an additional neutron or two may upset the binding energy and cause the atom to become unstable.

Actually not true.
It is true for a small although prominent part of cases.
Isotopes with equal number of protons and neutrons are stable for 9 out of the 10 first even elements (all except Be) and 4 first odd elements. Making 13 cases... out of 81 elements having any stable isotopes.
In most cases, elements like to have more neutrons than protons.
However, there is the conspicuous pattern here.
Small (and even at that!) nuclei like to have an equal number of protons and neutrons because it makes them more stable. Bigger nuclei (making up most of all nuclei) like to have more neutrons than protons.
This is because the electrostatic repulsion of protons favours neutrons over protons, for larger nuclei. Whereas for smaller nuclei, exchange repulsion (favouring equal numbers of neutrons and protons) prevails over the electrostatic repulsion (weaker in smaller nuclei).

 

[ChrisVer]

First of all you can't say that stable nuclei NEED equal neutrons ($N$) and protons ($Z$). This is an approximation, as you can see from looking at the chart of stable nuclei. It holds true for relatively low mass-nuclei, but as you go to heavier isotopes this diverses.

One thing that can help you understand the dynamics of the nuclei is the semi-empirical formula terms. Of course, this formula is not supposed to give you very exact answers. In this you have:
$E_B = a A - b A^{2/3} - c \frac{Z(Z-1)}{A^{1/3}}- d\frac{(N-Z)^2}{A}+\delta(N,Z)$
where $A$ is the atomic number and $Z$ the number of protons (defining the nucleus).

The first term is the "volume" term (reminder: the radius of a nuclei is scales with $A^{1/3}$, so the term appears as $radius^3$). That in general can give you an estimate of the binding energy between pairs of nucleons. The strong force is relatively a weak force at long distances, and so only nearest neighbours of nucleons can be considered to interact.

The second term is a "surface" term (it goes as $radius^2$). You can think of this term as a result of that the "outer nucleons" don't interact with the same number of nucleons as the ones inside the nuclei (they lack some nearest neighbours). Thus it is correcting what's missing in the first term.

The third term is the Coulomb force.

The fourth term is the "asymmetry" term. That's where the statement that nuclei prefer equal number of neutrons and protons stand. If they don't have that (i.e. $N\ne Z$) this term appears and makes the nucleus relatively unstable. The origin of this term is seen through Pauli's exclusion principle. In simple terms this tells us that identical fermions (such as protons or neutrons), cannot take the same quantum state. Thus you can view the nucleus as somekind of a container, divided in two, each part of which you can fill with neutrons and protons and they will occupy a given "position" in the container (actually the positions are energy levels). If now you start adding more nucleons (e.g. let's say neutrons), you will start filling your half container with higher and higher energy nucleons, while there are "lower" positions for protons to occupy. The imbalance you create is reflected in this term.

The last is the "pairing" term, no need to explain it right here.The important thing however to keep in mind, is that almost all these terms participate in determining the lowest energetic nucleus and thus the "stable" isotope. So, in order to maximize this energy, as we learned for school, we need to do something with the derivatives, and in particular the $\frac{dE_B}{dZ}=0$. That sometimes lead you in determining the ratio of neutrons-to-protons that result in stable nuclei. I won't do it now, but you end up with:
$\frac{N}{Z} \approx 1 + \frac{c}{2d} A^{2/3}$
which in general (for small $A$) results in the approximation $N \approx Z$, while for larger nuclei to win over the coulomb force you start needing to include more and more neutron rich nuclei (although the asymmetric term)
https://en.wikipedia.org/wiki/Neutron–proton_ratio

 

[snorkack]

First of all you can't say that stable nuclei NEED equal neutrons ($N$) and protons ($Z$). This is an approximation, as you can see from looking at the chart of stable nuclei. It holds true for relatively low mass-nuclei,

It´ s impossible even then. A nucleus with odd number of nucleons cannot have equal number of protons and neutrons.

The important thing however to keep in mind, is that almost all these terms participate in determining the lowest energetic nucleus and thus the "stable" isotope. So, in order to maximize this energy, as we learned for school, we need to do something with the derivatives, and in particular the $\frac{dE_B}{dZ}=0$. That sometimes lead you in determining the ratio of neutrons-to-protons that result in stable nuclei. I won't do it now, but you end up with:
$\frac{N}{Z} \approx 1 + \frac{c}{2d} A^{2/3}$
which in general (for small $A$) results in the approximation $N \approx Z$, while for larger nuclei to win over the coulomb force you start needing to include more and more neutron rich nuclei (although the asymmetric term)
https://en.wikipedia.org/wiki/Neutron–proton_ratio

And the reason why large nucleons become unstable has to do with balance between the surface term and the repulsion term.

Big nuclei have increasing Coulomb repulsion of protons. This can be only partly relaxed by neutron surplus because of the imbalance terms, so big nuclei still are more weakly bound. The smaller relative surface of big nuclei has decreasing effect.

This is why big nuclei become unstable.