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.

Answer 1

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.

\[ b=202-4+0 \]

\[ \therefore b=198 \]