- #1

JDiorio

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## Homework Statement

A patient with thyroid problems has radioactive iodine 123 deposited in his thyroid gland. If the iodine has an initial activity of 2.5 MBq, what is its activity after 24 days?

## Homework Equations

A(t)= Ao e^ -(ln(2)/half-life)*t

## The Attempt at a Solution

not sure what i did wrong here.. hope its not another calculating error.. but i found the half life of iodine 123 is 13.1 hours or .5458 days. So i have that ln(2)/.5458 times 24 = 30.48.

Then e^-30.48=5.795E-14.. i just multiply this number by 2.5E6 and get 1.45E-7 Bq.. not sure what I am doing wrong