A) 0.0019 mg
B) 1.019 mg
C) 1.109 mg
D) 0.019 mg
Correct Answer: D
Solution :
\[M={{M}_{0}}{{\left( \frac{1}{2} \right)}^{\frac{t}{{{T}_{1/2}}}}}\]\[=20\times {{\left( \frac{1}{2} \right)}^{\frac{3.6}{3.6}}}\]\[=20\times {{\left( \frac{1}{2} \right)}^{10}}=0.019\,mg\]
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