$A$ radioactive nucleus with $Z$ protons and $N$ neutrons emits an $\alpha$-particle,$2\beta^-$-particles,and $2$ gamma rays. The number of protons and neutrons in the nucleus left after the decay,respectively,are:

$A$ nucleus $X$ undergoes the following transformation:
$X \xrightarrow{\alpha} Y$
$Y \xrightarrow{2\beta} Z$
Then:

$A$ radioactive decay can form an isotope of the original nucleus with the emission of which particles?

Three $\alpha$-particles and one $\beta$-particle decay take place in series from an isotope $_{88}Ra^{238}$. Finally,the isotope obtained will be:

Assertion: ${}_{Z}{X^{A}}$ undergoes $2\alpha$-decays,$2\beta$-decays,and $2\gamma$-decays,and the daughter product is ${}_{Z-2}{X^{A-8}}$.
Reason: In $\alpha$-decays,the mass number decreases by $4$ and the atomic number decreases by $2$. In $\beta$-decays,the mass number remains unchanged,but the atomic number increases by $1$.

When a neutron is disintegrated to give a $\beta$-particle,which of the following is also emitted?