What is $\alpha  -$ , $\beta -$ and $\gamma -$decay ? Write its general formula. 

A plot of the number of neutrons $(N)$ against the number of protons ( $P$ )of stable nuclei exhibits upward deviation from linearity for atomic number, $Z>20$. For an unstable nucleus having $N / P$ ratio less than $1$ , the possible mode($s$) of decay is(are)

($A$) $\beta^{-}$-decay ( $\beta$ emission)

($B$) orbital or $K$-electron sasture

($C$) neutron emission

($D$) $\beta^{+}$-decay (positron emission)

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

Which can pass through $20 \,cm$ thickness of the steel

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

The isotope ${ }_5^{12} \mathrm{~B}$ having a mass $12.014 \mathrm{u}$ undergoes $\beta$-decay to ${ }_6^{12} \mathrm{C} .{ }_6^{12 .}$ has an excited state of the nucleus $\left({ }_6^{12} \mathrm{C}^*\right)$ at $4.041 \mathrm{MeV}$ above its ground state. If ${ }_5^{12} \mathrm{~F}$ decays to ${ }_6^{12} \mathrm{C}^*$, the maximum kinetic energy of the $\beta$-particle in units of $\mathrm{MeV}$ is ( $1 \mathrm{u}=931.5 \mathrm{MeV} / c^2$, where $c$ is the speed of light in vacuum).