Sunday

April 20, 2014

April 20, 2014

Posted by **Jammy** on Tuesday, October 5, 2010 at 4:46pm.

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

thanks!!

- chem -
**DrBob222**, Tuesday, October 5, 2010 at 4:49pmHere is how you do it.

You can do the math for the answer.

k = 0.693/t_{1/2}, 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 -
**Jammy**, Tuesday, October 5, 2010 at 4:52pmso 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 -
**DrBob222**, Tuesday, October 5, 2010 at 4:56pmYou 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 -
**Jammy**, Tuesday, October 5, 2010 at 5:08pmoh sorry! I didn't compute that part. lol. ill try that now

- chem -
**Jammy**, Tuesday, October 5, 2010 at 5:12pmi 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 -
**DrBob222**, Tuesday, October 5, 2010 at 5:49pmI 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 -
**Jammy**, Tuesday, October 5, 2010 at 5:58pmyes 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 -
**DrBob222**, Tuesday, October 5, 2010 at 6:10pmNo, 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**

log_{e}(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 e^{x}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 -
**Jammy**, Tuesday, October 5, 2010 at 6:24pmoh 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 -
**DrBob222**, Tuesday, October 5, 2010 at 7:37pmI 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

**Related Questions**

chemistry - Iodine-131 is used to destroy thyroid tissue in the treatment of an ...

Math please help! - Nuclear medicine technologists use the Iodine isotope I-131...

math-natural log - Nuclear medicine technologists use the Iodine isotope I-131, ...

chemistry - The disintegration of a radioisotope, such as I-131 to Xe-131, is ...

Chemistry - A viral contains radioactive iodine -131 with an activity of 2.0 mCi...

chemistry - The disintegration of a radioisotope, such as I-131 to Xe-131, is ...

science - the half-life of iodine-131 is 8 days. how much iodine-131 will be ...

Algebra - A defective nuclear reactor is leaking radioactive iodine-131. The ...

gen chem - The recommended dosage of iodine-131 is uCi/kg of body weight. How ...

Math - Please show me how to solve: Doctors can use radioactive iodine to treat ...