A) 0.0019 mg
B) 1.019 mg
C) 1.109 mg
D) 0.019 mg
Correct Answer: D
Solution :
Half life \[{{T}_{1/2}}=3.6\]days Initial quantity\[{{N}_{0}}=20\,mg\] Total time = 36 days The number of half lives \[n=\frac{t}{{{T}_{1/2}}}=\frac{36}{3.6}=10\] Hence, mass of radioactive substance left after 10 half lives \[N={{N}_{0}}\times {{\left( \frac{1}{2} \right)}^{n}}=20\times \frac{1}{1024}=0.019\,mg\]
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