$A$ nucleus $_Z^AX$ emits an $\alpha$-particle. The resultant nucleus then emits a ${\beta ^ + }$ particle. What will be the respective atomic number and mass number of the final nucleus?

$A$ radioactive decay forms an isotope of the original nucleus with the emission of which of the following particles?

$A$ radioactive nucleus emits a beta particle. The parent and daughter nuclei are

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 $1$.

The $Q$-value of the decay ${ }_{11}^{22} Na \rightarrow{ }_{10}^{22} Ne + e ^{+}+v$ is

Consider a radioactive nucleus $A$ which decays to a stable nucleus $C$ through the following sequence: $A \to B \to C$. Here $B$ is an intermediate nucleus which is also radioactive. Considering that there are $N_0$ atoms of $A$ initially,plot the graph showing the variation of the number of atoms of $A$ and $B$ versus time.