Match the column: Column I (Parent nucleus of the radioactive series) with Column II (Number of - an — Nuclear Chemistry and Radioactivity Chemistry Question
Question
Match the column: Column I (Parent nucleus of the radioactive series) with Column II (Number of $\alpha$- and $\beta$-particles emitted out to get the end product of that series)
💡 Solution & Explanation
(A) ${}_{92}^{235}\text{U}$ decays to ${}_{82}^{207}\text{Pb}$. $\Delta A = 28$, so $28/4 = 7\alpha$. $\Delta Z = 92 - 82 = 10$. $7(2) - x = 10 \implies x = 4\beta$ (A $\to$ S). (B) ${}_{92}^{238}\text{U}$ decays to ${}_{82}^{206}\text{Pb}$. $\Delta A = 32$, so $32/4 = 8\alpha$. $\Delta Z = 92 - 82 = 10$. $8(2) - x = 10 \implies x = 6\beta$ (B $\to$ R). (C) ${}_{94}^{241}\text{Pu}$ decays to ${}_{83}^{209}\text{Bi}$. $\Delta A = 32$, so $32/4 = 8\alpha$. $\Delta Z = 94 - 83 = 11$. $8(2) - x = 11 \implies x = 5\beta$ (C $\to$ Q). (D) ${}_{90}^{232}\text{Th}$ decays to ${}_{82}^{208}\text{Pb}$. $\Delta A = 24$, so $24/4 = 6\alpha$. $\Delta Z = 90 - 82 = 8$. $6(2) - x = 8 \implies x = 4\beta$ (D $\to$ P). Therefore, correct answer is A-S; B-R; C-Q; D-P.