An element $A$ decays into element $C$ by a two-step process:
$A \to B + {\;_2}He^4$
$B \to C + 2e^-$
Then:

$A$ nucleus with mass number $220$ initially at rest emits an $\alpha$-particle. If the $Q$ value of the reaction is $5.5\, MeV$,calculate the kinetic energy of the $\alpha$-particle in $MeV$.

$A$ positron is emitted from ${}^{23}Na_{11}$. The ratio of the atomic mass and atomic number of the resulting nuclide is

$A$ nucleus $X$ emits a beta particle to produce a nucleus $Y$. If their atomic masses are $M_{x}$ and $M_{y}$ respectively, the maximum energy of the beta particle emitted is (where $m_{e}$ is the mass of an electron and $c$ is the velocity of light):

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

$A$ radioactive nucleus ${}_{Z}^{A}X$ undergoes spontaneous decay in the sequence ${}_{Z}^{A}X \rightarrow {}_{Z-1}B \rightarrow {}_{Z-3}C \rightarrow {}_{Z-2}D$,where $Z$ is the atomic number of element $X$. The possible decay particles in the sequence are: