A) \[2T\]
B) \[1/2\,(\log \,2)\,T\]
C) \[4T\]
D) \[2\,(\log \,2)\,T\]
Correct Answer: D
Solution :
Given 75% decay 25% is left under decayed. This require a time. \[t=2{{T}_{1/2}}\] Where, \[{{T}_{1/2}}=\frac{\log 2}{\lambda }\] Also, \[T=\frac{1}{\lambda }\] \[t=\frac{2\log \,2}{\lambda }\] \[\Rightarrow \] \[t=2\,(\log 2)T\]
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