Answer
The radioactive decay is $\beta^{-} $ decay.
The answer is $(b)$
Work Step by Step
As we know the atomic numbers of $Te$ and $I$ are $52$ and $53$ respectively, let us represent the decay reaction by
${^{131}_{52}Te}\rightarrow{^{131}_{53}I}+X$
where $X$ is the decay product along with $I$
Here the mass number ($131$) of $Te$ and $I$ are not changing, but the atomic number increases from $52$ to $53$. This indicates that the product nuclei $I$ have gained a positive charge.
If the reaction was an alpha decay, then there must be a change in both atomic and mass numbers by 2 and 4 units respectively. A $\beta^{+} $ decay would decrease the atomic number by 1 unit and there won't be any change to the atomic number for a gamma decay.
Therefore, the reaction must be a $\beta^{-} $ decay where a neutron inside the nucleus is converted to a proton and an electron. Then $X$ must be a $\beta^{-} $ particle.
