Question

If ${}_{92}^{238}U$ emits $8$ $\alpha$-particles and $6$ $\beta$-particles, then the resulting nucleus is

• A. ${}_{82}^{206}U$
• B. ${}_{82}^{206}Pb$
• C. ${}_{82}^{210}Pb$
• D. ${}_{82}^{214}Pb$

Answer 1

Answer: (B)

${}_{82}^{206}Pb$

After emitting 1 $\alpha$ particle, the atomic number decreases by 2 and mass number decreases by 4.

After emitting 1 $\beta$ particle, the atomic number increases by 1 and mass number is unaffected.

So, in this situation, the mass number of the daughter nucleus is $238-8\times 4=206$ and the atomic number is $92-8\times 2+6=82$.

Thus, the answer is ${}_{82}^{206}Pb$.