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I apologize in advance for the stupid question, but I was wondering about neutron decay and elements.

Taking for example one of the most simple elements, Helium. Helium is made by 2 neutrons and 2 protons. The mean life of a neutron is $885$ seconds, so how is it possible for Helium, and other elements, to exist and being stable?

Related question

Why does the neutron decay, whilst the proton does not (or better: it seems it has a mean life ways greater than the life of the Universe)? Basically they are made by quarks (and gluons), with the difference $P = (u, u, d)$ and $N = (u, d, d)$.

Thank you very much!

Les Adieux
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Let me answer your second question first. A free neutron has a larger mass than a free proton (on the order of 1 MeV). An electron's mass is about half an MeV so it is energetically possible (meaning total energy is conserved) for a neutron to decay into a proton and an electron and still have enough energy left over to form an anti-neutrino (moving almost at the speed of light) and some kinetic energy for the electron as well. The free proton simply can't do the same because there is not enough energy in its mass ($mc^2$) to form a neutron (let alone a positron and neutrino as well). If it does decay (a big if) it must do so by an entirely different mechanism than the neutron. Since the proton is the lightest baryon, that decay would violate baryon number conservation (a very important part of our best theory).

Now let's consider your first question (stability of atoms). When neutrons and protons are bound inside a nucleus their effective masses are lowered but not always by the same amount. This lowering of the masses sometimes yields situations where it is simply not energetically possible for the decay to happen. For example, when the effective mass difference between a proton and a neutron is less than half an MeV there isn't enough energy to form either an electron or a positron and decay cannot occur. These situations are the stable elements (or isotopes as I am about to explain).

Isotopes result when we add neutrons to a nucleus but keep the number of protons the same. For isotopes that have too many neutrons the mass difference (neutron minus proton mass) can creep up enough that $\beta^-$ decay is possible. If too few neutrons are present the mass difference can flip its sign (proton minus neutron mass now positive) and $\beta^+$ decay becomes possible. For low mass elements the most stable situation is for nuclei to have neutron and proton numbers roughly equal. Because protons are charged and repel each other via the electromagnetic force whereas neutrons are neutral this situation changes as we move to more massive isotopes. Here the most stable isotopes are those with more neutrons than protons. When we get to the element Tc (43 protons) we encounter the lightest element with no stable isotopes (its most stable form has 55 neutrons but still decays with a half life of millions of years). As we go even higher the proton repulsion becomes more dominant and eventually at U (with 92 protons the last element with stable isotopes) the stable part of the periodic table is exhausted.

Lewis Miller
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  • The most stable form of Tc has 55 neutrons and a half-life of about 4.2 million years. Source: https://en.wikipedia.org/wiki/Isotopes_of_technetium#Technetium-98 – untreated_paramediensis_karnik Jan 17 '16 at 21:21
  • @no_choice99 Thanks. I had checked on this several days ago and mis-remembered. I'll edit the answer. – Lewis Miller Jan 17 '16 at 21:33
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    For a more detailed discussion of the stability of isotopes, see my answer to this question: https://physics.stackexchange.com/questions/228183/is-this-a-correct-demonstration-for-why-elements-above-untriseptium-cannot-exist/228290#228290 – Lewis Miller Jan 17 '16 at 21:41
  • What is the "effective mass" of a fundamental particle? Neutrons don't decay because there are no free states for the proton that are energetically possible to fill. – ProfRob Jan 17 '16 at 21:41
  • @RobJeffries I was trying to answer the question with language that is compatible with the question as asked. By "effective mass" I mean that if neutrons and protons retain their identities when bound in a nucleus (and they basically do because their quarks are still confined) their energies are reduced by their binding but as you say if there are no energetically allowed states for them to transition into there won't be a decay. I wanted to avoid talking about specific nuclear single particle states (what my PhD Thesis was about). That's why I addressed the free particle decay first. – Lewis Miller Jan 17 '16 at 21:59
  • Lewis, I like to think I understand binding energy. I know that the masses of a proton and a neutron bound as a deuteron are less than the masses of a free proton and a free neutron. But I've never heard of a situation wherein the mass of a bound neutron can be reduced to less than the mass of a bound prot0n in the same nucleus. Can you give me a reference for this mass-difference-causes-beta-decay idea? – John Duffield Jan 03 '17 at 13:57
  • I usually don't think in terms of masses for nucleons inside nuclei, but I used that language here because of the language used in the question. In actuality the sign of beta decay (whether an electron or positron results) depends on the binding energies of the neutron and proton orbitals involved. This in tern depends on the imbalance in neutron/proton ratio of the isotope. – Lewis Miller Jan 03 '17 at 15:20