Given the masses of various atomic particles $m _{ p }=1.0072 u , m _{ n }=1.0087 u$ $m _{ e }=0.000548 u , m _{ v }=0, m _{ d }=2.0141 u$ where $p \equiv$ proton, $n \equiv$ neutron, $e \equiv$ electron, $\overline{ v } \equiv$ antineutrino and $d \equiv$ deuteron. Which of the following process is allowed by momentum and energy conservation $?$

A nucleus of atomic mass $A$ and atomic number $Z$ emits ${M_1}$ particles. The atomic mass and atomic number of the resulting nucleus are

A nucleus of an element ${}_{84}{X^{202}}$ emits an $\alpha $-particle first, $\beta $ -particle next and then a gamma photon. The final nucleus formed has an atomic number

The total number of $\alpha$ and $\beta$ particles emitted in the nuclear reaction ${ }_{92}^{238} \mathrm{U} \rightarrow{ }_{82}^{214} \mathrm{~Pb}$ is

The radioactive decay of uranium into thorium is expressed by the equation $_{92}^{238}U \to _{90}^{234}Th + X,$ where $'X'$ is

An element $A$ decays into element $C$ by a two step process :

$A \to B + {\;_2}H{e^4}$

$B \to C + \;2{e^ - }$

Then