In the uranium radioactive series, the initial nucleus is ${ }_{92}^{238} \mathrm{U}$ and that the final…
In the uranium radioactive series, the initial nucleus is ${ }_{92}^{238} \mathrm{U}$ and that the final nucleus is ${ }_{82}^{206} \mathrm{~Pb}$. When uranium nucleus decays into lead, the number of $\alpha$-particles and $\beta$-particles emitted are
$4 \alpha, 5 \beta$
$5 \alpha, 3 \beta$
$6 \alpha, 7 \beta$
$8 \alpha, 6 \beta$
Solution
For $\alpha$ emission, the atomic number decreases by 2 and mass number decreases by 4.
$\therefore \quad$ No. of $\alpha$ particles emitted $=\frac{238-206}{4}=8$
$\therefore \quad$ New atomic number $=92-(8 \times 2)=76$
For $\beta$-emission, only the atomic number increases by 1.
$\therefore \quad$ No. of $\beta$-particles emitted $=82-76=6$