A) \[_{82}^{206}Pb\]
B) \[_{82}^{209}Pb\]
C) \[_{82}^{208}Pb\]
D) \[_{83}^{209}Br\]
Correct Answer: C
Solution :
New mass number \[{A}'=A-4{{n}_{\alpha }}\]\[=232-4\times 6=208\] atomic number\[{Z}'=Z+{{n}_{\beta }}-2{{n}_{\alpha }}\]\[=90+4-2\times 6=82\]
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