The energy spectrum of $\beta$-particles [number $N(E)$ as a function of $\beta$-energy $E$] emitted from a radioactive source is:

$A$ stationary nucleus of mass number $220$ emits an $\alpha$-particle. If the energy released in the process is $5.5 \, MeV$,find the kinetic energy of the $\alpha$-particle in $MeV$.

$A$ 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 $MeV$.

$A$ nucleus with $Z=92$ emits the following in a sequence: $\alpha, \alpha, \beta^{-}, \beta^{-}, \alpha, \alpha, \alpha, \alpha, \beta^{-}, \beta^{-}, \alpha, \beta^{+}, \beta^{+}$ and $\alpha$. The atomic number of the resulting nucleus is

An element $X$ decays into element $Z$ by a two-step process:
$X \rightarrow Y + 4e$
$Y \rightarrow Z + 2e^{-}$
Then:

In a radioactive decay chain reaction,${ }_{90}^{230} Th$ nucleus decays into ${ }_{84}^{214} Po$ nucleus. The ratio of the number of $\alpha$ to number of $\beta^{-}$ particles emitted in this process is. . . . .