Conservation of energy and the electron-degenerate pressure.
For the neutron to decay you must have
$$ n \to p + e^- + \bar{\nu}$$
or
$$ n + \nu \to p + e^- \quad. $$
In either case that electron is going to stay around, but in addition to the neutrons being in a degenerate gas, the few remaining electrons are also degenerate, which means that adding a new one requires giving it momentum above the Fermi surface and the energy is not available.
At first, consider two particles decay:
$A\rightarrow B + e^-$ Where A is initially rest.
So $\vec{p_B}+\vec{p_{e^-}}=0$
now
\begin{align}
\frac{p_B^2}{2m_B}+\sqrt{p_{e^-}^2+m_{e^-}^2}&=E_{released}
\\
\frac{p_{e^-}^2}{2m_B}+\sqrt{p_{e^-}^2+m_{e^-}^2}&=E_{released}
\tag{1}
\end{align}
see here you have uncoupled equation (equ.1) for $p_{e^-}$ .. So, solving above (equ.1) you will get a fixed $p_{e^-}$. Hence the energy($\sqrt{p_{e^-}^2+m_{e^-}^2}$) of the $\beta$ particle is always fixed in the two body $\beta$ decay. (you can find $p_{B}$ too using the momentum conservation formula)
At first, consider three particles decay:
$A\rightarrow B + e^-+\nu_e$ Where A is initially rest. So $\vec{p_B}+\vec{p_{e^-}}+\vec{\nu_e}=0$
so
\begin{align}
\frac{p_B^2}{2m_B}+\sqrt{p_{e^-}^2+m_{e^-}^2}+p_{\nu_e}&=E_{released}
\\
\frac{(\vec{p_{e^-}}+\vec {p_{\nu_e}})^2}{2m_B}+\sqrt{p_{e^-}^2+m_{e^-}^2}+p_{\nu_e}&=E_{released}
\tag{2}
\end{align}
Now see in contrast to the two particles decay equation, here we have five unknowns $\big{(}p_{e^-},p_{\nu},p_{B},\theta\ (\rm the\ angle\ between\ \vec{p_{e^-}},\vec{p_{\nu}}),\phi\ (\rm the\ angle\ between\ \vec{p_{e^-}},\vec{p_{B}})\big)$ but four coupled equations *. so we can't solve them uniquely. That's also what happens physically. You will get different values of $p_{e^-},p_{\nu},p_{B},\theta,\phi$ satisfying the four coupled equations. Hence different $\beta$ particles will have different energy($\sqrt{p_{e^-}^2+m_{e^-}^2}$) maintaining the statistics of decay process. Hence the continuous spectra.
*The four coupled equations are equ.2 and three equations which we can get by taking Dot products of $\vec{p_{e^-}},\vec{p_{\nu}}\rm\ and\ \vec{p_{B}}$ with the momentum conservation($\vec{p_B}+\vec{p_{e^-}}+\vec{\nu_e}=0$
) and remember they lie in a plane so two angles ($\theta\rm\ and\ \phi$) are sufficient.
Best Answer
Lead 206 is not stable against alpha decay. I was surprised to find out that even the doubly-magic nucleus 208Pb is unstable: the Q value (energy released in the decay) in units of amu is $+5.549(4)\times10^{-4}$.
Alpha decay half-lives approximately follow the rule that the log of the half-life varies linearly with $Q^{-1/2}$. The physics underlying this is that the nucleus has to tunnel out through the Coulomb barrier, and the barrier penetration probability depends exponentially on the square root of the amount by which the energy falls short of the classically allowed energy. (It's counterintuitive but true that the electrical force, which is repulsive, hinders these decays.) The $Q$ values for the "stable" isotopes of lead are extremely small. Therefore their half-lives are expected to be extremely long.
I guess this sort of make sense with hindsight, if you think about how alpha decay rates depend on $Z$. The lower you go in atomic number, the longer the half-lives become, until you reach lead, which is where we start to get effectively "stable" nuclei. It makes sense that as you get to lead, which is the boundary, there should be nuclei that have very long half-lives.
Keep in mind that lots of forms of matter can be theoretically unstable, but stable for practical purposes. For example, it's theoretically possible for 124Te to decay into two 62Ni nuclei plus four electrons and four antineutrinos, because the $Q$ value for such a decay is $+0.04$ amu. That doesn't mean that we ever expect to observe such a decay. Quantum mechanics says that whatever is not forbidden is compulsory, but that doesn't mean that it has a high probability.
That doesn't make much sense to me. Could you point us to a source on this? I imagine it's stable against beta decay, since it's even-even and close to the line of stability. The two processes are in any case independent. Having an additional decay process available will lower the overall half-life, not lengthen it.