(c) $\frac{3}{4}th$ active decay takes place to time $t = 2\left( T_{1/2} \right) \Rightarrow \frac{3}{4} = 2\left( T_{1/2} \right) \Rightarrow T_{1/2} = \frac{3}{8}s$
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.