Question

Which element belongs to $4n$ series?

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

Answer 1

Answer: (C)

$Pb^{208}$

Answer 2

Series of an element is decided by dividing its atomic mass by $4$. In case of $4\, n$ series, the remainder is zero. This is true only in case of $Pb ^{208}$. Thus, $Pb ^{208}$ belongs to $4\, n$ series.