Question

The last product of $4n$ series is:

• A. ${}_{82}^{208}Pb$
• B. ${}_{82}^{207}Pb$
• C. ${}_{82}^{209}Pb$
• D. ${}_{83}^{210}Bi$

Answer 1

The correct option is A ${}_{82}^{208}Pb$

4n series is the _${}_{90}^{232}Th$ decay series.
${}_{90}^{232}Th\stackrel{\alpha }{\to }{}_{88}^{228}Ra⟶....{⟶}_{84}^{212}{P}_o\stackrel{\alpha }{\to }{}_{82}^{208}{P}_o$
option (A) is correct.