The total number of $\alpha$ and $\beta$ particles emitted in the nuclear reaction ${ }_{92}^{238} U \rightarrow{ }_{82}^{214} Pb$ 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?

Find the $Q$-value and the kinetic energy of the emitted $\alpha$-particle in the $\alpha$-decay of $(a) \; ^{226}_{88} Ra$ and $(b) \; ^{220}_{86} Rn$.
Given:
$m(^{226}_{88} Ra) = 226.02540 \; u$
$m(^{222}_{86} Rn) = 222.01750 \; u$
$m(^{220}_{86} Rn) = 220.01137 \; u$
$m(^{216}_{84} Po) = 216.00189 \; u$
$m(^{4}_{2} He) = 4.002603 \; u$

Sometimes a radioactive nucleus decays into a nucleus which itself is radioactive. An example is
$^{38}S \xrightarrow{2.48 \ h} ^{38}Cl \xrightarrow{0.62 \ h} ^{38}Ar$
Assume that we start with $1000$ $^{38}S$ nuclei at time $t = 0$. The number of $^{38}Cl$ nuclei is zero at $t = 0$ and will again be zero at $t = \infty$. At what value of $t$ would the number of $^{38}Cl$ nuclei be a maximum?

After the decay of a single $\beta$ particle, the parent and daughter nuclei are

In the following nuclear reaction,$D \xrightarrow{\alpha} D_{1} \xrightarrow{\beta^-} D_{2} \xrightarrow{\alpha} D_{3} \xrightarrow{\gamma} D_{4}$. The mass number of $D$ is $182$ and the atomic number is $74$. The mass number and atomic number of $D_{4}$ respectively will be: