Consider a $\beta$ decay reaction:
${}_1^3H \to {}_2^3He + {e^{ - 1}} + \bar v$
The atomic masses of ${}_1^3H$ and ${}_2^3He$ are $3.016050 \, u$ and $3.016030 \, u$,respectively. Find the maximum possible energy of the electron in $MeV$.

In a radioactive decay, neither the atomic number nor the mass number changes. Which of the following would be emitted in the decay process?

$A$ radioactive nucleus decays as follows:
$X \xrightarrow{\alpha} X_1 \xrightarrow{\beta} X_2 \xrightarrow{\alpha} X_3 \xrightarrow{\gamma} X_4$
If the mass number and atomic number of $X_4$ are $172$ and $69$ respectively, then the atomic number and mass number of $X$ are:

Consider the following radioactive decay process:
${ }_{84}^{218} A \stackrel{\alpha}{\longrightarrow} A_1 \stackrel{\beta^{-}}{\longrightarrow} A_2 \stackrel{\gamma}{\longrightarrow} A_3 \stackrel{\alpha}{\longrightarrow} A_4 \stackrel{\beta^{+}}{\longrightarrow} A_5 \stackrel{\gamma}{\longrightarrow} A_6$
The mass number and the atomic number of $A_6$ are given by:

$A$ free neutron decays spontaneously into :

$_{86}A^{222} \to _{84}B^{210}$. In this reaction,how many $\alpha$ and $\beta$ particles are emitted?