JEE Advanced · 2016 · 22 May 2016 · Paper 1 · Chemical Kinetics

For $N/P<1$, the nucleus is proton-rich and needs to increase its neutron-to-proton ratio to reach stability. ${\beta }^{+}$-decay (positron emission) converts a proton to a neutron: $p\to n+e^{+}+{\nu }_e$, increasing $N/P$. Electron capture also converts a proton to a neutron: $p+e^{-}\to n+{\nu }_e$, increasing $N/P$. Both (B) and (D) are correct. ${\beta }^{-}$-decay would decrease $N/P$ (wrong direction), and neutron emission would also decrease $N/P$.