A) 0.91 mg
B) 0.25 mg
C) 0.5 mg
D) 0.125 mg
Correct Answer: D
Solution :
\[N={{N}_{0}}{{\left( \frac{1}{2} \right)}^{n}}\Rightarrow N={{N}_{0}}{{\left( \frac{1}{2} \right)}^{t/{{T}_{1/2}}}}\] \[\Rightarrow N=1\times {{\left( \frac{1}{2} \right)}^{\frac{8.1}{2.7}}}={{\left( \frac{1}{2} \right)}^{3}}=\frac{1}{8}\]\[\Rightarrow N=\frac{1}{8}mg=0.125\ mg\]
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