JEE Main 2010 — Nuclear Physics Question with Solution

In positive beta decay a proton is transformed into a neutron and a positron is emitted. $${\mathrm{p}}^{+}⟶{\mathrm{n}}^0+{\mathrm{e}}^{+}$$ no. of neutrons initially was $A-Z$ no. of neutrons after decay $(A-Z)-3\times 2$ (due to alpha particles) $+2\times 1$ (due to positive beta decay) The no. of proton will reduce by 8 . [as $3\times 2$ (due to alpha particles) $+2$ (due to positive beta decay)] Hence atomic number reduces by 8 .