Which of the following is in the increasing order for penetrating power

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 $2.$

 A nucleus  $_n{X^m}$ emits one $\alpha$ particle and two $\beta$ particles. The resulting nucleus is

Assertion : Radioactive nuclei emit ${\beta ^ - }$ particles.

Reason : Electrons exist inside the nucleus

${}^{238}U$ has $92$ protons and $238$ nucleons. It decays by emitting an Alpha particle and becomes

A radioactive nucleus undergoes a series of decay according to the scheme

$A\xrightarrow{\alpha }{{A}_{1}}\xrightarrow{\beta }{{A}_{2}}\xrightarrow{\alpha }{{A}_{3}}\xrightarrow{\gamma }{{A}_{4}}$

If the mass number and atomic number of $A$ are $180$ and $72$ respectively, then what are these number for $A_4$