A nuclear decay is possible if the mass of the parent nucleus exceeds the total mass of the decay particles. If $M(A, Z)$ denotes the mass of a single neutral atom of an element with mass number $A$ and atomic number $Z$, then the minimal condition that the $\beta$ decay $X_Z^A \rightarrow Y_{Z+1}^A+\beta^{-}+\bar{v}_e$ will occur is ( $m_e$ denotes the mass of the $\beta$ particle and the neutrino mass $m_v$ can be neglected)

In a radioactive decay chain, the initial nucleus is ${}_{90}^{232}Th$.  At the end there are $6\,\,\alpha -$ particles and $4\,\,\beta -$ particles with are emitted. If the end nucleus is ${}_Z^AX\,,\,A$ and $Z$ are given by

${ }_{92}^{238} U$ atom disintegrates to ${ }_{84}^{214} Po$ with a half of $45 \times 10^9$ years by emitting $\operatorname{six} \alpha-$ particles and $n$ electrons. Here, $n$ is

In the nuclear reaction $_{85}{X^{297}} \to Y + 4\alpha ,\;Y$ is

A radioactive nucleus undergoes $\alpha$- emission to form a stable element. What will be the recoil velocity of the daughter nucleus if $V$ is the velocity of $\alpha$-emission and $A$ is the atomic mass of radioactive nucleus

A nucleus $X$ undergoes following transformation

$X \stackrel{a}{\longrightarrow} Y$

$Y \underset{2 \beta}{\longrightarrow} Z$

then