A) \[_{82}^{208}Pb\]
B) \[_{82}^{207}Pb\]
C) \[_{82}^{209}Pb\]
D) None of these
Correct Answer: A
Solution :
Series End product \[4n\] \[_{82}^{208}Pb\] \[4n+1\] \[_{83}^{209}Bi\] \[4n+2\] \[_{82}^{206}Pb\] \[4n+3\] \[_{82}^{207}Pb\]
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