Question

Match List I with List II

List I	List II
(A) Rn22286 → Po21884	(III) α particle
(B) Bi21483 → Po21484	(I) β− particle
(C) Th23490 → U23492	(I) β− particle
(D) Na2211 → Na2210	(II) β+ particle

• A. A-II, B-I, C-IV, D-III
• B. A-III, B-II, C-IV, D-I
• C. A-I, B-II, C-IV, D-II
• D. A-I, B-II, C-III, D-IV

Hint:

Alpha decay results in the emission of a \( \alpha \)-particle, while beta decay results in the emission of a \( \beta \)-particle.

Answer 1

The Correct Option is A

- A: \( ^{222}_{86}Rn \to ^{218}_{84}Po \) undergoes alpha decay, emitting a \( \alpha \)-particle, so the correct match is \( \text{A-II} \). - B: \( ^{214}_{83}Bi \to ^{214}_{84}Po \) undergoes beta decay, emitting a \( \beta \)-particle, so the correct match is \( \text{B-I} \). - C: \( ^{234}_{90}Th \to ^{234}_{92}U \) undergoes alpha decay, emitting a \( \alpha \)-particle, so the correct match is \( \text{C-IV} \). - D: \( ^{22}_{11}Na \to ^{22}_{10}Na \) undergoes beta decay, emitting a \( \beta \)-particle, so the correct match is \( \text{D-III} \). Thus, the correct matching is: A-II, B-I, C-IV, D-III.