My question is simply that - what makes a nucleus unstable? What exactly causes a nucleus to start breaking apart in the first place? Is it the Coulomb force between the neighboring protons?
I'm just curious.
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asked Nov 29, 2014 at 1:33
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$\begingroup$ Breaking apart in radioactive decay? $\endgroup$– HDE 226868Commented Nov 29, 2014 at 1:57
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$\begingroup$ By the way, I believe the strong nuclear force between the protons would definitely overwhelm any electromagnetic forces between them. Lubos Motl explains it pretty well here. $\endgroup$– HDE 226868Commented Nov 29, 2014 at 2:10
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$\begingroup$ Possible duplicates: physics.stackexchange.com/q/9661/2451 and links therein. $\endgroup$– Qmechanic ♦Commented Jul 19, 2016 at 17:44
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2 Answers
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It is instructive to look at chart of isotopes ,
number of neutrons on the x axis and protons on the y. The stable (black) isotopes diverge from the diagonal, more neutrons are needed to neutralize the coulomb repulsion of the protons, for stability.
The main forces are the coulomb force (repulsive) and the strong force (attractive) , but the specific energy levels will depend on quantum numbers . If there are too many neutrons, there exists a probability for the neutron decay to express itself , if too many protons a positron emission will bring the balance.
Instability means that there are lower energy levels in the collective nuclear potential of the specific number of protons and neutrons to which the system will stabilize. The decay time can take from centuries to nanoseconds depending on the probability of decay for the given system. A number of models have been developed to explain the specific behavior.
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Even if the strong nuclear force is the most powerful force over subatomic distances. The electrostatic force is almost always significant, and, in the case of beta decay (which also involves a transformation of a proton into a neutron), the weak nuclear force is also involved. But the basic answer is yes, you cannot put two protons together (Helium 2 doesnt exist, at least in a stable form), you need neutrons to compensate and add more strong force to avoid the electrostatic repulsion. Even if the strong force is stronger, it acts over very short distances, that is why elctrostatic repulsion wins sometimes, specially on large nuclei whare the strong force of a proton on the core has little influence on a proton on the external shell, but the electrostatic force, is still strong as it decays slowly with distance (remember that the range of the strong force is limited to distances on the order of a few $10^{-15}$ meters)
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2$\begingroup$ well, it is only partial as an answer, one has to consider the quantum mechanical aspect of the effective potential, stable energy levels for a certain combination of protons and neutrons and unstable ones ( in isotopes where the number of neutrons changes, and thus the nuclear force) whether there are energy levels where the binding energy is such that a fission will release energy. $\endgroup$– anna vCommented Nov 29, 2014 at 5:24
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$\begingroup$ @annav I agree, I tried to keep it simple, the OP is not a professional $\endgroup$– user65081Commented Nov 29, 2014 at 5:27