A nucleus $_Z^AX$ emits an $\alpha$- particle. The resultant nucleus emits a ${\beta ^ + }$ particle. The respective atomic and mass no. of the final nucleus will be

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 $2.$

For the given reaction, the particle $X$ is $_6{C^{11}}{ \to _5}{B^{11}} + {\beta ^ + } + X$

In the disintegration series

$_{92}^{238}U\xrightarrow{\alpha }x\xrightarrow{\beta }_Z^AY$

The value of $Z$ and $A$ respectively will be

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

Alpha particles are