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

Question

The half-life of a radioactive substance is 3.6. 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

Tardigrade Admin · Accepted Answer

Half life Ty2=3.6 days Initial quantity N0=20 mg Total time =36 days The number of half lives n=(t/T1 / 2)=(36/3.6)=10 Hence, mass of radioactive substance left after 10 half lives N=N0 ×((1/2))n=20 × (1/1024)=0.019 mg

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

0.0019 mg

1.019 mg

1.109 mg

0.019 mg

Half life $T_{y_{2}} = 3.6$ days
Initial quantity $N_{0} = 20 m g$
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
$N = N_{0} \times (\frac{1}{2})^{n} = 20 \times \frac{1}{1024} = 0.019 m g$

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