I The fate of neutron rich nuclei

  • Thread starter Geofleur
  • Start date

Geofleur

Science Advisor
Gold Member
423
176
I've been trying to understand why adding neutrons to a nucleus will eventually destabilize it; I would like to know if the following explanation is correct:

The neutron has a slightly higher mass than the proton. But higher mass translates into higher energy because ## E = mc^2 ##. However, systems tend to evolve toward lower energies, like a ball rolling down a hill and coming to rest in a depression.

A free neutron will decay (in about 10 min) into a proton, an electron, and an anti-electron neutrino. This reaction doesn't happen to neutrons inside stable atoms (see below). But in a neutron rich atom, the nucleus can transition to a lower state if a neutron turns into a proton, because the proton is lighter than the neutron.

The reason that all neutrons don't just spontaneously decay into protons is that both protons and neutrons are fermions, meaning that no more than one can occupy any given quantum state. After a point, a neutron decaying into a proton will result in a proton with high enough energy to compensate for the mass difference between the proton and neutron, because there are no low energy slots left for the newly formed proton to occupy.

What do you guys think?

Later edit: I found this post later on, which seems to agree with what I'm saying? https://www.physicsforums.com/threads/why-are-neutron-rich-isotopes-unstable.625062/
 
Last edited:
33,373
9,100
But in a neutron rich atom, the nucleus can transition to a lower state if a neutron turns into a proton, because the proton is lighter than the neutron.
The mass difference of the free particles is small, the available energy states in the nuclei are much more important. In proton-rich nuclei you have the opposite type of decay (beta+ instead of beta-).
 
1,363
135
The mass difference of the free particles is small, the available energy states in the nuclei are much more important. In proton-rich nuclei you have the opposite type of decay (beta+ instead of beta-).
Sometimes more important, yes.
Look at the simple symmetric nuclei:
n vs. p: n has decay energy of 782 keV, half-life 10 minutes
t vs. 3He: t has decay energy of 18 keV, half-life 12 years. Obviously neutron is bound more strongly to d than p - but that 764 keV difference is not enough to stabilize t. It does, however, cause 3He to be a strong neutron absorber, releasing these 764 keV in:
3He+n→t+p
5He vs. 5Li: both unbound
7Li vs. 7Be: 7Be is unstable, yet does NOT have β+ decay! The opposite reaction of β- is not β+ - it is electron capture. 7Be has that, with energy of 862 keV - and β+ is therefore impossible.
9Be vs. 9B: 9Be is bound by 1680 keV, 9B unbound
11B vs. 11C: 11C does have decay energy of about 1980 keV for electron capture, and therefore can also undergo β+ decay, with energy of about 960 keV, which turns out to be the main decay path, about 20 minutes half-life.
 

Geofleur

Science Advisor
Gold Member
423
176
What I understand from these answers is the following:

The neutron, left to itself, decays; even inside a nucleus it tends to decay if the nucleus is too neutron-rich. Protons, on the other hand, do not decay on their own (well, some speculative theories say they do, after a really, really long time). But inside the nucleus a proton can convert into a neutron through electron capture or positron emission. Since protons repel each other through the Coulomb interaction, electron-capture (or sometimes positron emission) is energetically favorable when the nucleus is proton-rich. So we have a balancing act between beta decay and electron capture (and sometimes positron emission).

Is that a good way to summarize the state of affairs?
 
Last edited:
33,373
9,100
@snorkack: Yes there are some exceptions where the proton/neutron mass difference matters, but I don't think they are important here.

Beta+ for proton-rich nuclei is really analogous to beta- for neutron-rich nuclei. Electron capture is an additional decay mechanism if there are electrons around, it is energetically favorable only because of the rest energy of the involved particles (using the electron gives up to 511 keV, not needing the positron gives another 511 keV difference).
 
1,363
135
Beta+ for proton-rich nuclei is really analogous to beta- for neutron-rich nuclei. Electron capture is an additional decay mechanism if there are electrons around, it is energetically favorable only because of the rest energy of the involved particles (using the electron gives up to 511 keV, not needing the positron gives another 511 keV difference).
I believe that viewing electron capture as the proper inverse of beta decay is the more useful.
This is the basis of Mattauch rule (violated only by 123Te and 180mTa).
When you consider two isobars different by only one proton or neutron, when electrons are not around, many of them (starting with 7Be) do not have the energy to produce an electron-positron pair.
But when electrons are around, since the rest mass of neutrinos, if any, is small, then either of the isobars is higher in energy and might decay into the others.
 

Want to reply to this thread?

"The fate of neutron rich nuclei" You must log in or register to reply here.

Related Threads for: The fate of neutron rich nuclei

Replies
2
Views
1K
Replies
10
Views
6K
  • Posted
Replies
18
Views
4K
Replies
2
Views
2K
  • Posted
Replies
3
Views
5K

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top