Match List I with List II
| LIST I | LIST II | ||
|---|---|---|---|
| A. | \( {}^{208}_{84}\mathrm{Po}\rightarrow{}^{204}_{82}\mathrm{Pb}+\ldots \) | I. | Electron |
| B. | \( {}^{32}_{15}\mathrm{P}\rightarrow{}^{32}_{16}\mathrm{S}+\ldots \) | II. | Neutron |
| C. | \( {}^{14}_{7}\mathrm{N}\rightarrow{}^{14}_{6}\mathrm{C}+\ldots \) | III. | Alpha particle |
| D. | \( {}^{2}_{1}\mathrm{H}+{}^{3}_{1}\mathrm{H}\rightarrow{}^{4}_{2}\mathrm{He}+\ldots \) | IV. | Positron |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → A-III, B-I, C-IV, D-II
In every nuclear reaction the mass number (top) and the charge number (bottom) balance on both sides. The missing particle carries the difference.
(A) \( {}^{208}_{84}\mathrm{Po}\rightarrow{}^{204}_{82}\mathrm{Pb}+\ldots \) — this matches (III) Alpha particle. Missing: A = 4, Z = 2, that is \( {}^{4}_{2}\mathrm{He} \).
(B) \( {}^{32}_{15}\mathrm{P}\rightarrow{}^{32}_{16}\mathrm{S}+\ldots \) — this matches (I) Electron. Missing: A = 0, Z = −1 (beta-minus decay).
(C) \( {}^{14}_{7}\mathrm{N}\rightarrow{}^{14}_{6}\mathrm{C}+\ldots \) — this matches (IV) Positron. Missing: A = 0, Z = +1.
(D) \( {}^{2}_{1}\mathrm{H}+{}^{3}_{1}\mathrm{H}\rightarrow{}^{4}_{2}\mathrm{He}+\ldots \) — this matches (II) Neutron. Missing: A = 5 − 4 = 1, Z = 2 − 2 = 0.