$\beta$-decay means emission of electron from

$A$ stationary nucleus of mass number $220$ emits an $\alpha$-particle. If the energy released in the process is $5.5 \, MeV$,find the kinetic energy of the $\alpha$-particle in $MeV$.

In a radioactive decay chain,the initial nucleus is ${}_{90}^{232}Th$. At the end,$6$ $\alpha$-particles and $4$ $\beta$-particles are emitted. If the final nucleus is ${}_{Z}^{A}X$,then $A$ and $Z$ are given by:

Statement-$1$: $A$ nucleus having energy $E_{1}$ decays by $\beta^{-}$ emission to a daughter nucleus having energy $E_{2}$,but the $\beta^{-}$ rays are emitted with a continuous energy spectrum having an end-point energy $E_{1} - E_{2}$.
Statement-$2$: To conserve energy and momentum in $\beta$ decay,at least three particles must take part in the transformation.

After one $\alpha$ and two $\beta$ emissions,what happens to the nucleus?

$37$ Rutherford equals