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.
My understanding of the Q-factor is that it's the difference in binding energies before and after the [nuclear] reaction.
This is wrong. The Q-value is the difference between the rest masses of the initial state and the final state. The binding energy subtracts out the particle masses. Because the neutron mass is different from the proton mass, if your data source gives you just binding energies, you will be led astray.
(Beware that the “Q-factor” of a damped resonator is different from the “Q-value” of a nuclear decay.)
Could someone please provide […] which of the above calculations is correct?
To see which calculation is correct, get some data and try to match it. Two systems where you might make progress are the decay of the free neutron (where the initial and final binding energies are both zero) and the decay of tritium to helium-3.
As I frequently advise people, I prefer the “mass excess” over the raw mass, the binding energy, or other equivalent data. The mass excess subtracts off the constant 1 a.m.u. per nucleon, and is usually already tabulated in energy units, so you don’t have to carry around four or five significant figures to get two or three significant figures in a mega-eV decay.
Best Answer
The notation (incoming, outgoing) is for scattering processes, not for spontaneous decays.