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
$A = 208;\,\,Z = 80$
$A = 202;\,\,Z = 80$
$A = 208;\,\,Z = 82$
$A = 200;\,\,Z = 81$
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:
${ }_1^1 H$ | $1.007825 u$ | ${ }_2^1 H$ | $2.014102 u$ | ${ }_3^1 H$ | $3.016050 u$ | ${ }_2^4 He$ | $4.002603 u$ |
${ }_3^6 Li$ | $6.015123 u$ | ${ }_7^3 Li$ | $7.016004 u$ | ${ }_70^30 Zn$ | $69.925325 u$ | ${ }_{34}^{82} Se$ | $81.916709 u$ |
${ }_{64}^{152} Gd$ | $151.919803 u$ | ${ }_{206}^{82} Gd$ | $205.974455 u$ | ${ }_{209}^{83} Bi$ | $208.980388 u$ | ${ }_{84}^{210} Po$ | $209.982876 u$ |
$1.$ 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