A) \[6.5%\]
B) \[12.5%\]
C) \[25.5%\]
D) \[33.3%\]
Correct Answer: B
Solution :
Amount of radioactive material left is \[N={{N}_{0}}{{\left( \frac{1}{2} \right)}^{n}}\] Where n = number of half-lives. \[=3\] \[{{N}_{0}}\]= original amount of the radioactive material. \[\Rightarrow \] \[N={{N}_{0}}{{\left( \frac{1}{2} \right)}^{3}}=\frac{{{N}_{0}}}{8}\] \[=\frac{100}{8}%\] Of \[{{N}_{0}}=125%\] of \[{{N}_{0}}\]
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