A) Bi-209
B) Rn-222
C) Th-230
D) Pb-207
Correct Answer: A
Solution :
Neptunium \[({}_{93}N{{p}^{237}})\] is a member of (n+1) series, hence, its end product will be Bi-209 \[(209=4\times 52+1)\].You need to login to perform this action.
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