An element $A$ emits an $\alpha$-particle and forms $B$. $A$ and $B$ are

$X$ $\xrightarrow{-\alpha } Y$ $\xrightarrow{-\beta } Z$ $\xrightarrow{-\beta } W$
In the above sequence of reaction,the elements which are isotopes of each other are

Two nuclei which are not identical but have the same number of nucleons represent

Relative abundance (in percentage) of ${}^{14}C$ isotope is

$_{35}X^{88}$ is an unstable isotope,which decays in two successive steps to produce the stable isotope $_{32}Z^{84}$ as $_{35}X^{88}$ $\xrightarrow{I} Y$ $\xrightarrow{II} _{32}Z^{84}$. The correct statement is (possible emissions are $\alpha, \beta^-$,positron,neutron,and $_1H^1$,and in each step only one particle is involved).

In the following radioactive transformation given below $R$ $\xrightarrow{\alpha} X$ $\xrightarrow{\beta} Y$ $\xrightarrow{\beta} Z$; the nuclei $R$ and $Z$ are