An element $X$ decays into element $Z$ by a two-step process:
$X \rightarrow Y + 4e$
$Y \rightarrow Z + 2e^{-}$
Then:

Alpha rays emitted from a radioactive substance are

Which of the following can penetrate through an aluminum sheet of $20 \, cm$ thickness?

$A$ nucleus with $Z=92$ emits the following in a sequence: $\alpha, \alpha, \beta^{-}, \beta^{-}, \alpha, \alpha, \alpha, \alpha, \beta^{-}, \beta^{-}, \alpha, \beta^{+}, \beta^{+}$ and $\alpha$. The atomic number of the resulting nucleus is

$U^{238}$ 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?

$A$ 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?