The isotope of _90Ra^231 can be converted to _90Th^227 by — Nuclear Chemistry and Radioactivity Chemistry Question

💡 Solution & Explanation

Step 1 - Conservation of Mass Number Let $N_\alpha$ = alpha particles emitted, $N_\beta$ = beta particles emitted. Each $\alpha$ reduces mass number by 4; $\beta$ has mass number 0. $$231 = 227 + 4N_\alpha$$ $$4N_\alpha = 4 \implies N_\alpha = 1$$ Step 2 - Conservation of Atomic Number Each $\alpha$ reduces $Z$ by 2; each $\beta^-$ increases $Z$ by 1. $$90 = 90 + 2(1) - N_\beta$$ $$N_\beta = 92 - 90 = 2$$ Step 3 - Balanced Nuclear Reaction $$\ce{^{231}_{90}Ra -> ^{227}_{90}Th + ^4_2He + 2^{0}_{-1}e}$$ Step 4 - Elimination of Wrong Options - (A) 1 $\alpha$ only: mass number OK (227) but $Z = 88 \ne 90$ — incorrect - (B) 4 $\beta$: mass number unchanged at 231 — incorrect - (C) 2$\alpha$ + 2$\beta$: mass number would be $231 - 8 = 223 \ne 227$ — incorrect - (D) 1$\alpha$ + 2$\beta$: $A = 227$ ✓ and $Z = 90$ ✓ $$\boxed{D}$$