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Question
A nucleus of an element has the symbol 84P202, and emits an α-particle and then a β-particle. The final nucleus is bQa Find a and b.
Solution
(i) On the emission of an alpha `(He_2^4)` particle:
\[\ce{_84^202P->[α]_82^19X(new nucleus)}\]
(ii) On the emission of a beta particle:
\[\ce{_82^198X->[β-particle]_83^198Q(new nucleus)}\]
`"_83^198Q` is the same as `"_b^aQ`. Hence a = 198, b = 83.
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