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Question
A nucleus $${}_{84}\mathrm{X}^{202}$$ of an element emits an alpha particle followed by a beta particle. The final nucleus is $${}_{\mathrm{a}}\mathrm{Y}^{\mathrm{b}}$$. Find a and b.
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Solution
Given: Initial nucleus = \({}_{84}^{202}X\)
Final nucleus = \({}_{a}^{b}Y\)
Number of \(\alpha\)-particles emitted = 1
Number of \(\beta\)-particles emitted = 1
The atomic number (a).
Each \(\alpha\)-particle emission decreases the atomic number by 2, while each \(\beta^{-}\)-particle emission increases it by 1.
\[ a=84-2+1 \]
\[ a=82+1 \]
\[ \therefore a=83 \]
The mass number (b).
Each \(\alpha\)-particle emission decreases the mass number by 4, while \(\beta\)-particle emission does not change it.
\[ b=202-4+0 \]
\[ \therefore b=198 \]
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