In the given nuclear reaction, the element $X$ is :

${ }_{11}^{22} Na \rightarrow X + e ^{+}+v$

Write nuclear reaction equations for

$(i)$ $\alpha$ -decay of $^{226}_{88} Ra$

$(ii)$ $\alpha$ -decay of $_{94}^{242} Pu$

$(iii)$ $\beta$ -decay of $_{15}^{32} P$

$(iv)$ $\beta$ -decay of $^{210}_{83}Bi$

$(v)$ $\beta^{+}$ -decay of $_{6}^{11} C$

$(vi)$ $\beta^{+}$ -decay of $_{43}^{97} Tc$

$(vii)$ Electron capture of $^{120}_{54} Xe$

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 $\beta$-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 $2.$

Actinium $231$, ${}^{231}A{c_{89}}$, emit in succession two $\beta  - $particles, four alphas, one $\beta  $ and one alpha plus several $\gamma $ rays. What is the resultant isotope

Assertion : The ionising power of $\beta  - $ particle is less compared to $\alpha  - $ particles but their penetrating power is more.

Reason : The mass of $\beta  - $ particle is less than the mass of $\alpha  - $ particles