Question

In the Uranium radioactive series, the initial nucleus is ${}_{92}^{238}U$ and final nucleus is ${}_{82}^{206}Pb$. When the Uranium nucleus decays to Lead, the number of a particles emitted is ___ and the number of $\beta$ particles emitted is_________

• A. 6, 8
• B. 8, 6
• C. 16, 6
• D. 32, 2

Answer 1

Answer: (B)

According to question,
Initial nuclei $={}_{92}^{238}U$
Final nuclei $={}_{82}^{206}Pb$
Change in atomic number $=92-82=10$
When an $\alpha$ - particle emits from a nuclei, then its atomic number is
decreased by $2$ units and atomic mass is decreased by $4$ units.
When a $\beta$-particle emits from a nuclei, then atomic number of parent nuclei is increased by $1$ unit whereas its atomic mass remains same.
Change in atomic mass $=238-206=32$
Since, atomic mass changes only due to emission $\alpha$-particle, hence number of emitted $\alpha$-particle
$=\frac{32}{4}=8$
Number of emitted $\beta$-particles = total change in atomic number due to emission of $\alpha$-particles total change in atomic number $=8\times 2-10=6$