$A$ radioactive element $X$ emits six $\alpha$-particles and four $\beta$-particles,leading to the final product ${ }_{82}^{208} Pb$. The element $X$ is:

$A$ free neutron decays spontaneously into:

If $Alpha$,$Beta$,and $\Gamma$ rays carry the same momentum,which has the longest wavelength?

The $A_{92}U^{238}$ nucleus decays to a $Pb^{214}_{82}$ nucleus. The number of $\alpha$ and $\beta^{-}$ particles emitted are:

The intensity of gamma radiation emitted from a source is $I_0$. After passing through a lead sheet of thickness $36 \, mm$,its intensity becomes $I_0/8$. Find the thickness of the lead sheet in $mm$ when the intensity becomes $I_0/2$.

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 two-body decay process,it was predicted theoretically that the kinetic 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.,$n \rightarrow p + e^- + \bar{\nu}_e$,around $1930$,Pauli explained the observed electron energy spectrum. Assuming the anti-neutrino $(\bar{\nu}_e)$ to be massless and possessing negligible energy,and the neutron to be at rest,momentum and energy conservation principles are applied. From this calculation,the maximum kinetic energy of the electron 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 for question $1$ and $2$.