A) \[_{89}A{{c}^{225}}\]
B) \[_{90}T{{h}^{232}}\]
C) \[_{15}{{P}^{35}}\]
D) \[_{92}{{U}^{235}}\]
Correct Answer: D
Solution :
The series \[4n+3\] is known as actinium series. It shows that if mass number is divided by 4 then the remainder must be 3. Hence, the member of this series is\[_{92}{{U}^{235}}\]. \[\because \] \[\frac{235}{4}=58+\operatorname{Remainder}\,\,3\]
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