Answer:
The correct answer would be = 0.1593 mCi.
Explanation:
The initial 123I = 2.55 mCi (given)
We know that the half-life of radioactive I-123 = 12 hours approximately.
So, we can say that in 48 hours there will be four half-lives of 123I,
Hence after four half-life or 48 hours, the final 123I would be:
= 2.55 * (1/2⁴)
= 2.55 * (1/16)
= 0.1593 mCi
Thus, the correct answer would be = 0.1593 mCi.
