A nucleus with $\mathrm{Z}=92$ emits the following in a sequence: $\alpha, \beta^{-}, \beta^{-} \alpha,…
A nucleus with $\mathrm{Z}=92$ emits the following in a sequence:
$\alpha, \beta^{-}, \beta^{-} \alpha, \alpha, \alpha, \alpha, \alpha, \beta^{-}, \beta^{-}, \alpha, \beta^{+}, \beta^{+}, \alpha$. Then $\mathrm{Z}$ of the resulting nucleus is
$76$
$78$
$82$
$74$
Solution
No. of $\alpha$ particles emitted $=8, \quad$ No. of $\beta^{-}$particles emitted $=4, \quad$ No.of $\beta^{+}$particles emitted $=2$
$z=92-2 \times 8+4-2=78$