The end product of the decay of $ { }_{90} Th^{232} $ is $ { }_{82} Pb^{208} $. The number of $ \alpha $ and $ \beta $ particles emitted are respectively:

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?

In the given radioactive decay,the number of $\alpha$ and $\beta$ particles emitted are respectively: $_{90}X^{200} \rightarrow _{80}Y^{168}$

In a radioactive decay, neither the atomic number nor the mass number changes. Which of the following would be emitted in the decay process?

$37$ Rutherford equals

${ }_{92} U^{235}$ undergoes successive disintegrations with the end product of ${ }_{82} Pb^{203}$. The number of $\alpha$ and $\beta$ particles emitted are