# chem

posted by on .

Iodine-131 is often used in nuclear medicine to obtain images of the thyroid. If you start with 3.6 109 I-131 atoms, how many are left after approximately 1 month? I-131 has a half-life of 8.0 days.

can you tell me how to do that problem and what answer you got?

thanks!!

• chem - ,

Here is how you do it.
You can do the math for the answer.
k = 0.693/t1/2, then substitute k into the equation below.

ln(No/N) = kt
No is the starting number.
N is the ending number.
t is the time. If you use the half-life in days you must convert 1 month to day to use that for t.

• chem - ,

so i converted 8 days to one month and got.266666...

and i plugged that in for t 1/2 so when i divided those two numbers i got 2.59875 but that doesnt seem nearly correct at all

what am i doing wrong?

• chem - ,

You aren't doing anything wrong. So far so good. Why did you stop there? My instructions were to calculate k (which you did), then substitute that into the equation below that and solve for N.

• chem - ,

oh sorry! I didn't compute that part. lol. ill try that now

• chem - ,

i did those calculations and got 1385281385. so i put that for my answer and it said it was wrong. So, now i only have one more entry left.

• chem - ,

I don't know what you're doing wrong but I don't get that at all. You may be using months for half life and days for t BUT
ln(3.6 x 10^9/N) = 2.59875(1)
Is this what you had? If so, solve that and show how you did it and I'll find the error.

• chem - ,

yes that is wut i had. so you end up dividing 3.6e9 by 2.59875 and you get 1385281385 which is wut i originaly had. but its wrong.. wait.. i searched on google for the equation so i can see it layed out and this website im on says

nt/no for the first part of the equation..? is that what it is suppose to be?

• chem - ,

No, that isn't what you have and it isn't what you do.
ln(3.6 x 10^9/N) = 2.59875(1)
ln(3.6 x 10^9/N) = 2.59875
Note that the left is an ln term so it is
loge(3.6 x 10^9/N) = 2.59875
So you take the antilog of both sides. The antilog of ln(3.6 x 10^9/N) is just 3.6 x 10^9/N. To take the antilog of 2.59875 you punch the ex key on your calculator and obtain 13.4469 so
(3.6 x 10^9/N) = 13.4469 and
N = (3.6 x 10^9/13.4469) = 2.something x 10^8 or close to that but you report only two significant figures, not all of the extra digits you get from the calculator. I think that rounds to 2.7 x 10^8 atoms.

• chem - ,

oh man i totaly forgot the log!! oops!

I tried that and it said it was wrong. that was my last submission. But its ok. i have 13 other questions so if i get those i will at least get an A-.

but.. i just want to know for future reference why that is wrong. Do you know?

• chem - ,

I can point out two or three things. The first is that the problem says "how many are left after APPROXIMATELY one month. We really don't know what is meant by approximately so that is poor wording. It could mean 30 days or it could mean 31 days.
If there are 30 days in the month, then
k = 0.693/8 = 0.08663 and
ln(3.6 x 10^9/N) = 0.08663 x 30 days and N = 2.7 x 10^8 atoms.

If we make it 31 days,
k = same and
ln(3.6 x 10^9/N) = 0.08663 x 31
and N = 2.455 x 10^9 which I would round to 2.5 x 10^9.

If we do it the other way,
8 days in a 30 day month = 8/30 = 0.2667 and k = 2.59875, then
ln(3.6 x 10^9/N) = 2.59875*1
and N = the same 2.7 x 10^8 atoms.

or for 31 days, then
8/31 = xx and k = 2.685
Then ln(3.6 x 10^9/N) = 2.685*1
and N = the same 2.455 which I would round to 2.5 x 10^8 atoms.

On a related topic, that equation you found on google, if it is Nt/No, (the one I typed is No/Nt) that will = -kt and both are seen in the literature. Both are correct.
ln(No/N) = kt or ln(Nt/No) = -kt