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

💡 Solution & Explanation

Step 1 - Decay Particle Rules Each emitted particle changes mass number $A$ and atomic number $Z$: | Particle | Symbol | $\Delta A$ | $\Delta Z$ | |----------|--------|-----------|-----------| | Alpha | $\ce{^4_2He}$ | $-4$ | $-2$ | | Beta-minus | $\ce{^0_{-1}e}$ | $0$ | $+1$ | Let $x$ = number of $\alpha$ emissions, $y$ = number of $\beta^-$ emissions. Step 2 - Conservation of Mass Number $$A_\text{initial} = A_\text{final} + 4x$$ $$231 = 227 + 4x$$ $$4x = 4 \implies \boxed{x = 1}$$ Step 3 - Conservation of Atomic Number $$Z_\text{initial} = Z_\text{final} + 2x - y$$ $$90 = 90 + 2(1) - y$$ $$y = 2 \implies \boxed{y = 2}$$ Step 4 - Verify the Decay Chain $$\ce{^{231}_{90}Th ->[\alpha] ^{227}_{88}Ra ->[\beta^-] ->[\beta^-] ^{227}_{90}Th}$$ - After 1 $\alpha$: $Z = 88,\ A = 227$ (Ra-227 intermediate) - After 2 $\beta^-$: $Z = 90,\ A = 227$ ✓ = Th-227 Step 5 - Evaluate Options - **(A) One $\alpha$ emission**: gives $Z = 88$, not 90. Product is Ra-227, not Th-227. Incorrect. - **(B) Four $\beta$ emissions**: mass unchanged at 231, not 227. Incorrect. - **(C) Two $\alpha$ + two $\beta$**: mass drops by 8 to 223, not 227. Incorrect. - **(D) One $\alpha$ + two $\beta$**: $A = 227$, $Z = 90$. **Correct.** $$\boxed{\text{Answer: D}}$$