Question

If by successive disintegration of ${{}_{92}U}^{238}$, the final product obtained is ${{}_{82}Pb}^{206}$, then how many number of $\alpha$ and $\beta$ particles are emitted?

• A. $8$ and $6$
• B. $6$ and $8$
• C. $12$ and $6$
• D. $8$ and $12$

Answer 1

The correct option is A $8$ and $6$

We know that atomic number of the parent nucleus decreases by $2$ and mass number decreases by $4$ in each alpha decay whereas atomic number of the parent nucleus increases by $1$ while mass number remains the same in each beta minus decay.
Change in mass number $\Delta A=238-206=32$
Number of alpha particles $N_{\alpha }=\frac{32}{4}=8$
Decrease in mass number due to alpha decays only $=2\times 8=16$
Net decrease in mass number $\Delta Z=92-82=10$
So increase in mass number due to beta particles $=16-10=6$
Number of beta particles $N_{\beta }=\frac{6}{1}=6$