The nucleus ${ }_{88}^{226} Ra$ is converted into ${ }_{82}^{206} Pb$ by the emission of alpha $(\alpha)$ and beta $(\beta)$ particles. The number of alpha and beta particles emitted are respectively:

Statement-$1$: $A$ nucleus having energy $E_{1}$ decays by $\beta^{-}$ emission to a daughter nucleus having energy $E_{2}$,but the $\beta^{-}$ rays are emitted with a continuous energy spectrum having an end-point energy $E_{1} - E_{2}$.
Statement-$2$: To conserve energy and momentum in $\beta$ decay,at least three particles must take part in the transformation.

$A$ radioactive decay chain starts from $_{93}Np^{237}$ and produces $_{90}Th^{229}$ by successive emissions. The emitted particles can be:

$A$ radioactive nucleus with $Z$ protons and $N$ neutrons emits an $\alpha$-particle,$2\beta^-$-particles,and $2$ gamma rays. The number of protons and neutrons in the nucleus left after the decay,respectively,are:

In the following equation representing $\beta^{-}$ decay,the number of neutrons in the nucleus $X$ is ${ }_{83}^{210} Bi \longrightarrow X + { }_{-1}^{0} e + \bar{\nu}$

The equation $_{Z}X^{A} \to _{Z+1}Y^{A} + _{-1}e^{0} + \bar{\nu}$ represents: