${}^{238}U$ has $92$ protons and $238$ nucleons. It decays by emitting an alpha particle and becomes:

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 $1$.

$U^{238}$ decays into $Th^{234}$ by the emission of an $\alpha$-particle. There follows a chain of further radioactive decays,either by $\alpha$-decay or by $\beta$-decay. Eventually,a stable nuclide is reached,and after that,no further radioactive decay is possible. Which of the following stable nuclides is the end product of the $U^{238}$ radioactive decay chain?

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

Out of the following, which one is not emitted by a natural radioactive substance?

In the given reaction $_Z{X^A} \to _{Z+1}{Y^A} \to _{Z-1}{K^{A-4}} \to _{Z-1}{K^{A-4}}$,radioactive radiations are emitted in the sequence: