JEE Advanced 2023 Paper 1 · Q14 · Radioactivity
List-I shows different radioactive decay processes and List-II provides possible emitted particles. Match each entry in List-I with an appropriate entry from List-II. List-I: (P) $^{238}_{92}U \to {}^{234}_{91}Pa$ (Q) $^{214}_{82}Pb \to {}^{210}_{82}Pb$ (R) $^{210}_{81}Tl \to {}^{206}_{82}Pb$ (S) $^{228}_{91}Pa \to {}^{224}_{88}Ra$ List-II: (1) one $\alpha$ particle and one $\beta^+$ particle, (2) three $\beta^-$ particles and one $\alpha$ particle, (3) two $\beta^-$ particles and one $\alpha$ particle, (4) one $\alpha$ particle and one $\beta^-$ particle, (5) one $\alpha$ particle and two $\beta^+$ particles
Reveal answer + step-by-step solution
Correct answer:A
Solution
For each decay, balance mass number $A$ (changes only via $\alpha$, $\Delta A=4$) and charge $Z$ (\$\alpha$: $-2$; $\beta^-$: $+1$; $\beta^+$: $-1$).
(P) ${}^{238}_{92}U\to{}^{234}_{91}Pa$: $\Delta A=4\Rightarrow 1\alpha$ (leaves $Z=90$); to reach $91$ need $+1\Rightarrow 1\beta^-$. Match (4).
(Q) ${}^{214}_{82}Pb\to{}^{210}_{82}Pb$: $\Delta A=4\Rightarrow 1\alpha$ (leaves $Z=80$); to reach $82$ need $+2\Rightarrow 2\beta^-$. Match (3).
(R) ${}^{210}_{81}Tl\to{}^{206}_{82}Pb$: $\Delta A=4\Rightarrow 1\alpha$ (leaves $Z=79$); to reach $82$ need $+3\Rightarrow 3\beta^-$. Match (2).
(S) ${}^{228}_{91}Pa\to{}^{224}_{88}Ra$: $\Delta A=4\Rightarrow 1\alpha$ (leaves $Z=89$); to reach $88$ need $-1\Rightarrow 1\beta^+$. Match (1).
Therefore $P\to 4,\ Q\to 3,\ R\to 2,\ S\to 1$.
Want to drill deeper? Open this question inside SolveKar AI and tap "Why?" on any step — the AI Mentor reads your full solution context and explains until it clicks. Download SolveKar AI →