A) \[T{{h}^{232}}\]
B) \[N{{p}^{237}}\]
C) \[A{{c}^{227}}\]
D) \[{{U}^{238}}\]
Correct Answer: C
Solution :
\[P{{b}^{207}}(207=4\times 51+3)\]belongs to 4n + 3 series or Actinium \[(A{{c}^{227}})\] series. It starts from \[{{U}^{235}}.\]
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