Question

A nucleus of an element has the symbol ₈₄P²⁰², and emits an α-particle and then a β-particle. The final nucleus is _bQ^a Find a and b.

Answer 1

(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.