Question

In the reaction ${}_{15}{p}^{30}{\to }_{14}s{i}^{30}$, The change requires the emission of

• A. $\alpha$ - particle
• B. $\beta$ -particle
• C. neutron
• D. positron

Answer 1

The correct option is D positron

Let the emitted particle is X.
Mass number on both side are equal, so mass number of X is 0 units.
Atomic number on L.H.S. is 1 unit greater than R.H.S.
So, element X has 1 unit of atomic number.
Therefore, element X has configuration of
${}_1^{0}X{=}_1^{0}e$ i.e. positron