$A$ nucleus with $Z = 92$ emits $\alpha, \alpha, \beta^-, \beta^-, \alpha, \alpha, \alpha, \alpha, \beta^-, \beta^-, \alpha, \beta^+, \beta^+, \alpha$ particles in sequence. What is the $Z$ of the resulting nucleus?

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

An $U^{238}$ nucleus is initially at rest. It decays by emitting an $\alpha$-particle with speed $u$. Find the recoil speed of the daughter nucleus.

$U^{238}$ 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?

In the disintegration series
$_{92}^{238}U \xrightarrow{\alpha} X \xrightarrow{\beta} _{Z}^{A}Y$
The values of $Z$ and $A$ respectively will be:

Some radioactive nucleus may emit