# Nucleus stability: Neutron & Proton Ratio

by quinteboy
Tags: neutron, nucleus, proton, ratio, stability
 P: 2 Hi I'm looking for a good description (high-school to 1st year level) why the ratio of protons to neutrons matters for stability. Single neutrons are unstable, but in a nucleus they are stable and the more nucleons you have the greater the strong force. So why can't you have stable isotopes say uranium where excess neutrons increase the binding energy per nucleon ratio to a point to make the atom stable?
 Sci Advisor P: 5,768 http://en.wikipedia.org/wiki/Isotope Try the above. It is a complex subject that cannot be easily answered.
 PF Patron P: 10,391 Here is my possibly incorrect understanding: A neutron is made up of 1 up quark and 2 down quarks. Down quarks are slightly more massive than up quarks. Matter in general wants to be in the lowest energy state it can be in. Normally, when the neutron is in free space (Not in a nucleus), there is no reason for it to stay in that slightly more energetic form, therefore it decays into a proton, which is made up of 2 up quarks and 1 down quark, in order to reach that lower energy state. (As far as we know Protons are completly stable and never decay at normal temps/energies) Now, an atomic nucleus requires neutrons to provide extra Strong Force to hold protons together against their repulsive positive charges. Staying together inside a nucleus means that the protons and neutrons have a lower combined energy than their combined individual energies if they were alone by themselves. Once you start adding neutrons to a nucleus, you reach a certain point, which varies with each different element, that the energy of the nucleus exceeds that breakeven limit and the nucleus now wants to do something to reach a lower level of energy. It does this by decaying by the various methods to form another element. One of these is the Beta decay method, which is also the exact method that a free neutron decays into a proton while by itself in space. Hope this helps.
P: 2

## Nucleus stability: Neutron & Proton Ratio

Thanks for your responses, I think I understand (enough) what's happening
 P: 918 Yea, Drakkith's reasoning appears correct.
 PF Patron Sci Advisor Emeritus P: 5,306 Drakkith's argument doesn't seem correct to me, because it never invokes the exclusion principle, and I don't think you can explain this without the exclusion principle. Also, it invokes the mass difference between up and down quarks, which I don't think is necessary in order to understand why the line of stability has the general shape it does. (The low-mass line of stability is N=Z, so clearly the asymmetry in mass doesn't matter there.) First let's consider the case without electrical interactions, which is relevant for low masses. If you add more and more neutrons, you have to put them in states more and more energy, because the low-energy states are full. Same if you add more and more protons. At any given point as you're filling the states, your best option is to add a neutron if the lowest empty state is a neutron state, a proton if the lowest is a proton state. For heavy nuclei, the line of stability curves away from N=Z. This is because of the electrical interaction, which is unfavorable to adding more protons.
PF Patron
P: 10,391
 Quote by bcrowell Drakkith's argument doesn't seem correct to me, because it never invokes the exclusion principle, and I don't think you can explain this without the exclusion principle. Also, it invokes the mass difference between up and down quarks, which I don't think is necessary in order to understand why the line of stability has the general shape it does. (The low-mass line of stability is N=Z, so clearly the asymmetry in mass doesn't matter there.) First let's consider the case without electrical interactions, which is relevant for low masses. If you add more and more neutrons, you have to put them in states more and more energy, because the low-energy states are full. Same if you add more and more protons. At any given point as you're filling the states, your best option is to add a neutron if the lowest empty state is a neutron state, a proton if the lowest is a proton state. For heavy nuclei, the line of stability curves away from N=Z. This is because of the electrical interaction, which is unfavorable to adding more protons.
That's pretty much what my post was saying, but not in so elequant language. =)

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