Question

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

• A. 0.0019 mg
• B. 1.019 mg
• C. 1.109 mg
• D. 0.019 mg

Answer 1

Answer: (D)

0.019 mg

Answer 2

Half life $T_{1 / 2}=3.6$ days Initial quantity $N_{0}=20\, mg$ Total line $=36$ 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 $N =N_{0} \times\left(\frac{1}{2}\right)^{n}$ $=20 \times \frac{1}{1024}=0.019 \,mg$