Alpha decay is treated as a special case and is usually not included in the term "fission", although it arguably is a fission (especially in the case of light elements where the alpha can represent an appreciable fraction of the original mass).
So, for the purposes of a quiz you want "fission".
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
The electrons freed from the bounds of the fissioned nucleus will follow conservation of momentum and will move according to the kinetic energy they have. What will happen to them will depend on the medium they are in. Their kinetic energy will be too high for them to meet up with the fragments constructively, so there is very low probability the new nuclei will tie up with the freed electrons. The nuclei will pick up electrons from the medium, or remain ions.
Generally the electrons will lose energy by scattering on the fields of the atoms and molecules of the medium until they are captured. If in a neutral gas they may be captured by an atom/molecule and turn it into a negatively charged ion. If in solid , metal particularly they will end up in the conduction band.