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

In a radioactive decay, neither the atomic number nor the mass number changes. Which of the following would be emitted in the decay process

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 $2.$

$_{90}^{232}Th$ an isotope of thorium decays in ten stages emitting six  $\alpha$-particles and four $\beta$-particles in all. The end product of the decay is

A deutron is bombarded on $_8{O^{16}}$ nucleus and $\alpha$-particle is emitted. The product nucleus is

In the nuclear decay given below 

$_z{X^A}{ \to _{z + 1}}{Y^A}{ \to _{z - 1}}{K^{A - 4}}{ \to _{z - 1}}{K^{A - 4}}$

the particles emitted in the sequence are