Question

The element ${}_{90}^{232}Th$ belongs to thorium series. Which of the following will act as the end product of the series?

• A. ${}_{82}^{208}Pb$
• B. ${}_{82}^{209}Bi$
• C. ${}_{82}^{206}Pb$
• D. ${}_{82}^{207}Pb$

Answer 1

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

4n series of ${}_{90}^{232}Th$ ends with ${}_{82}^{208}Ps.$
So option A is correct.