# How come neutrons in a nucleus don't decay?

I know outside a nucleus, neutrons are unstable and they have half life of about 15 minutes. But when they are together with protons inside the nucleus, they are stable. How does that happen?

I got this from wikipedia:

When bound inside of a nucleus, the instability of a single neutron to beta decay is balanced against the instability that would be acquired by the nucleus as a whole if an additional proton were to participate in repulsive interactions with the other protons that are already present in the nucleus. As such, although free neutrons are unstable, bound neutrons are not necessarily so. The same reasoning explains why protons, which are stable in empty space, may transform into neutrons when bound inside of a nucleus.

But I don't think I get what that really means. What happens inside the nucleus that makes neutrons stable?

Is it the same thing that happens inside a neutron star's core? Because, neutrons seem to be stable in there too.

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Why do You accept the first (and in this case wrong) answer? –  Georg Apr 25 '11 at 19:05
The second answer was more helpful to me but, the first one was concise. I don't understand why the first answer got down voted. Probably something I'm missing. Anyway, I'll accept the second one, just to be safe. But please someone help me understand why the first one is incorrect. –  Aria Apr 30 '11 at 15:41

Spontaneous processes such as neutron decay require that the final state is lower in energy than the initial state. In (stable) nuclei, this is not the case, because the energy you gain from the neutron decay is lower than the energy it costs you to have an additional proton in the core.

For neutron decay in the nuclei to be energetically favorable, the energy gained by the decay must be larger than the energy cost of adding that proton. This generally happens in neutron-rich isotopes:

An example is the $\beta^-$-decay of Cesium: $$\phantom{Cs}^{137}_{55} Cs \rightarrow \phantom{Ba}^{137}_{56}Ba + e^- + \bar{\nu}_e$$

For a first impression of the energies involved, you can consult the semi-empirical Bethe-Weizsäcker formula which lets you plug in the number of protons and neutrons and tells you the binding energy of the nucleus. By comparing the energies of two nuclei related via the $\beta^-$-decay you can tell whether or not this process should be possible.

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I think you could improve this answer by adding some examples after the figure, referencing the figure. This would help the answer have a more homogenous "level". –  Bjorn Wesen Apr 25 '11 at 18:52
I dont see how saying "it would be more unstable, if it happened, so it doesn't" is possible. What is it about the nucleus that stops the random process of decay? –  Jonathan. May 10 at 8:51
@Jonathan It's the Bethe-Weizsäckere formula that semi-empirically tells you what the binding energy of any particular nucleus is. Spontaneous processes only happen from states of high- to low-energy. –  Lagerbaer May 10 at 14:14
Ok, but I mean how does it "know" that the end state is higher or lower energy? Does the neutron have a lower energy when inside the nucleus so it just can't decay, or is it something like it decays but then immediately reverts? –  Jonathan. May 10 at 15:46
The Pauli Exclusion Principle states that no two identical fermions (neutrons and protons are fermions - they have half-integer spins and obey Fermi-Dirac statistics) can occupy the same quantum state at the same time. If the neutron were to $\beta$-decay as: $$n \longrightarrow p + e^- + \bar{\nu_e}$$ then this freshly minted proton will try to occupy the quantum state with the lowest possible energy. However, since there are already loads of protons in the nucleus, this 'new' proton can't do that, and so will be forced to occupy a state with higher energy. In order to get to that state, it must absorb some energy. This is why neutrons don't usually $\beta$-decay inside the nucleus. Do remember that $\beta$-decay of neutrons inside the nucleus isn't unheard of - just uncommon.