Answer:
The correct answer is - C. 0.500 mg.
Explanation:
In this case, the half-life of the iodine-131 is 8 days and the initial amount is given 4 mg. According to this after every 8 days, half of the initial value of the iodine-131 remains only.
8 days or 1st half-life = A(i)* 1/2
16 days or two half-life = A(i)* 1/4
24 days or three half-life = A(i)* 1/8
and the remaining amount A would be
= A(i)*1/2^n
= 4 * 1/2^3
= 4 *1/8
= 0.500 mg
