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Q. The half-life of a radioactive substance is $3.6$. How much of $20\,mg$ of this radioactive substance will remain after $36$ days ?

JIPMERJIPMER 2004Nuclei

Solution:

Half life $T_{y_{2}}=3.6$ 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
$N=N_{0} \times\left(\frac{1}{2}\right)^{n}=20 \times \frac{1}{1024}=0.019\, mg$