Question

When ${}_{15}{P}^{30}$ decays to become ${}_{14}S{i}^{30}$, which particle is released ?

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

Answer 1

The correct option is D positron

The nuclear reaction : ${}_{15}{P}^{30}\to$ $14S{i}^{30}+$ ${}_{+1}{e}^0$

Thus a positron is emitted during the decay of ${}_{15}{P}^{30}$ into ${}_{14}S{i}^{30}$.