In the nuclear reaction $_{92}U^{238} \to _{Z}Th^{A} + _{2}He^{4}$,the values of $A$ and $Z$ are:

When ${}_{92}U^{238}$ changes into ${}_{82}Pb^{206}$,the number of $\alpha$ and $\beta^-$ particles emitted are:

The ${ }_{92}^{235} U$ atom disintegrates to ${ }_{82}^{207} Pb$ with a half-life of $10^9 \ yr$. In the process,it emits $7 \ \alpha$ particles and $n \ \beta^{-}$ particles. Here,$n$ is

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?

$A$ radioactive element ${ }_{92}^{242} X$ emits two $\alpha$ particles,one electron and two positrons. The product nucleus is represented by ${ }_{P}^{234} Y$. The value of $P$ is

$A$ radioactive nucleus of mass $M$ emits a photon of frequency $v$ and the nucleus recoils. The recoil energy will be