1. Dec 4, 2009

### JDiorio

1. The problem statement, all variables and given/known data
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?

2. Relevant equations

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

3. 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 im doing wrong

2. Dec 5, 2009

### ideasrule

1.45E-7 Bq is right, but are you sure the question says 24 days and not 24 hours? A thyroid scan is usually done 24 hours after the patient ingests a pill containing iodine-123, or 30 minutes after iodine is intravenously injected.

3. Dec 5, 2009

### JDiorio

yeah i copied this directly from my hw site.. hmm.. its telling me that it is the wrong answer.. not sure what to do.. but it definitely says 24 days..

4. Dec 5, 2009

### JDiorio

so is this answer correct?? because when i typed it in .. it said it wasn't..

5. Dec 5, 2009

### JDiorio

I don't wanna keep bumping this thread.. but i really need to know what im suppose to do.. i have one more attempt at this problem on my hw site and i need all the points i can get.. I just need to know what is wrong with this answer

6. Dec 5, 2009

### denverdoc

have you tried the answer for 24 hours?

7. Dec 7, 2009

### cynthiayyf

I just found out that ur I123 half-life was wrong b/c I was working on the same HW. Try using 0.55day as the half-life, and you will get the correct answer!!

8. Dec 7, 2009

### JDiorio

THANK YOU!! yeah i just did it with that number. So stupid. I use 13.2 hrs and it was 13.1. thanks a lot.