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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?
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The correct answer is:
0.019 mg
Half life $T_{1 / 2}=3.6$ days
Initial quantity $N_0=20 \mathrm{mg}$
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
$N=N_0 \times\left(\frac{1}{2}\right)^n=20 \times \frac{1}{1024}=0.019 \mathrm{mg}$
Initial quantity $N_0=20 \mathrm{mg}$
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
$N=N_0 \times\left(\frac{1}{2}\right)^n=20 \times \frac{1}{1024}=0.019 \mathrm{mg}$
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