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
You have been misled. An alpha particle decay is exactly that - the unstable nucleus emits an alpha particle, which is a He nucleus (2 protons and 2 neutrons). Radioactive decay is not an "atomic" process.
The emitted alpha particles (He nuclei) have a +2 positive charge. They leave behind a negatively charged daughter atom (a di-anion). The di-anion also won't be around long; it recoils from the alpha particle emission and will interact with its surroundings, passing on the spare electrons. It is partly this process that releases heat in radioactive materials.
The alpha particles do not go very far unless they are in a vacuum. In air, they travel typically a few cm before interacting with other atoms and picking up a couple of electrons to become He atoms.