The transformations of protons into neutrons and vice versa are _____ -order reactions. — Nuclear Chemistry and Radioactivity Chemistry Question

💡 Solution & Explanation

Step 1 - Nuclear Transformations Involved Inside an unstable nucleus, protons can transform into neutrons via $\beta^+$ emission or electron capture, and neutrons can transform into protons via $\beta^-$ emission: $$\text{Proton} \to \text{Neutron} + e^+ + \nu_e \quad (\beta^+\ \text{decay})$$ $$\text{Neutron} \to \text{Proton} + e^- + \bar{\nu}_e \quad (\beta^-\ \text{decay})$$ Step 2 - Kinetics of Radioactive Decay For any radioactive process, the rate of transformation depends only on the number of unstable nuclei present at that instant: $$\frac{dN}{dt} = -\lambda N$$ This is mathematically a **first-order rate law**: rate $\propto N^1$ (the first power of the number of nuclei). The integrated form confirms first-order: $$N = N_0 e^{-\lambda t}$$ Step 3 - Evaluate Options - **(A) zero order**: Zero-order rate would be constant ($-dN/dt = \text{const}$), independent of N. This is not observed for radioactive decay. Incorrect. - **(B) first order**: The decay rate is proportional to $N^1$. All radioactive processes, including proton-neutron interconversions, are first-order reactions. **Correct.** - **(C) second order**: Would require rate $\propto N^2$, implying two-body collisions. Nuclear decay is a spontaneous single-nucleus process. Incorrect. - **(D) half order**: No physical basis for half-order nuclear decay. Incorrect. $$\boxed{\text{Answer: B — First-order reactions}}$$