Radioactivity was discovered by

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 $1$.

$A$ nucleus $_n{X^m}$ emits one $\alpha$ particle and two $\beta$ particles. The resulting nucleus is

Explain $\beta$-decay and how a radioactive nucleus emits $\beta$-particles,even if there are no $\beta$-particles in the nucleus. Why does the mass number of the radioactive nuclide not change during $\beta$-emission?

$A$ radioactive substance emits

$A$ nucleus $_n{X^m}$ emits one $\alpha$ and one $\beta$ particle. The resulting nucleus is: