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

The electron emitted in beta radiation originates from

A nucleus decays by ${\beta ^ + }$ emission followed by a gamma emission. If the atomic and mass numbers of the parent nucleus are $Z$ and $A$ respectively, the corresponding numbers for the daughter nucleus are respectively.

The mass of a nucleus ${ }_Z^A X$ is less that the sum of the masses of $(A-Z)$ number of neutrons and $Z$ number of protons in the nucleus. The energy equivalent to the corresponding mass difference is known as the binding energy of the nucleus. A heavy nucleus of mass $M$ can break into two light nuclei of masses $m_1$ and $m_2$ only if $\left(m_1+m_2\right)M^{\prime}$. The masses of some neutral atoms are given in the table below:

$1.$ The correct statement is:

$(A)$ The nucleus ${ }_3^6 Li$ can emit an alpha particle

$(B)$ The nucleus ${ }_{84}^{210} P_0$ can emit a proton

$(C)$ Deuteron and alpha particle can undergo complete fusion.

$(D)$ The nuclei ${ }_{30}^{70} Zn$ and ${ }_{34}^{82} Se$ can undergo complete fusion.

$2.$ The kinetic energy (in $keV$ ) of the alpha particle, when the nucleus ${ }_{84}^{210} P _0$ at rest undergoes alpha decay, is:

$(A)$ $5319$ $(B)$ $5422$ $(C)$ $5707$ $(D)$ $5818$

Give the answer question $1$ and $2.$

The $\beta$-decay process, discovered around $1900$ , is basically the decay of a neutron ( $n$ ), In the laboratory, a proton ( $p$ ) and an electron ( $e ^{-}$) are observed as the decay products of the neutron. therefore, considering the decay of a neutron as a tro-body dcay process, it was predicted theoretically that thekinetic energy of the electron should be a constant. But experimentally, it was observed that the electron kinetic energy has a continuous spectrum. Considering a three-body decay process, i.e. $n \rightarrow p+ e ^{-}+\bar{v}_{ e }$, around $1930,$ Pauli explained the observed electron energy spectrum. Assuming the anti-neutrino $\left(\bar{v}_{ e }\right)$ to be massless and possessing negligible energy, and neutron to be at rest, momentum and energy conservation principles are applied. From this calculation, the maximum kinetic energy of the lectron is $0.8 \times 10^6 eV$. The kinetic energy carried by the proton is only the recoil energy.

$1.$ What is the maximum energy of the anti-neutrino?

$(A)$ Zero

$(B)$ Much less than $0.8 \times 10^6 \ eV$

$(C)$ Nearly $0.8 \times 10^6 \ eV$

$(D)$ Much larger than $0.8 \times 10^6 \ eV$

$2.$ If the anti-neutrino had a mass of $3 eV / c ^2$ (where $c$ is the speed of light) instead of zero mass, what should be the range of the kinetic energy, $K$, of the electron?

$(A)$ $0 \leq K \leq 0.8 \times 10^6 \ eV$

$(B)$ $3.0 eV \leq K \leq 0.8 \times 10^6 \ eV$

$(C)$ $3.0 eV \leq K < 0.8 \times 10^6 \ eV$

$(D)$ $0 \leq K < 0.8 \times 10^6 \ eV$

Give the answer question $1$ and $2.$

${ }_{92}^{238} U$ 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