# Determining mass of isotope

1. Dec 13, 2006

### FlipStyle1308

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

One of the many isotopes used in cancer treatment is 19879Au, with a half-life of 2.70 d. Determine the mass of this isotope that is required to give an activity of 200 Ci.

2. Relevant equations

200Ci x 3.7x10^10 decays/s = 7.4 x 10^12 decays/s
T1/2 = ln2/lambda = 0.693/lambda

3. The attempt at a solution

I looked through my book in the section on half-life and radioactive dating, but could not find an equation(s) that would seem useful for this problem, since it involves mass (which is introduced later on in the chapter). The only equation I found relevant is the one I listed in (b). Any help would be appreciated. Thanks!

Last edited: Dec 13, 2006
2. Dec 13, 2006

### OlderDan

The fundamental principle underlying exponential decay is that activity is proportional to the number of atoms of a radioactive substance available to decay. Surely you have an equation of the form

A = A_o*exp{-kt}

and one of the form

A = k'N where N is the number of atoms in the sample.

How are k and k' related to half life?

3. Dec 13, 2006

### FlipStyle1308

We don't have those equations in this book, but I see R=R_o_e^(-lambdat), and R=lambdaN. I'm assuming that your A is the same as my R, and your k is my lambda. If so, then I can say that lambda and lambda' is the decay constant, which gives the rate of decay in half-life.

I did some work while waiting for a response, so check if this is correct:

R=R_o_e^(-lambdat)
7.4x10^12 decays/s = (3.7x10^10 decays/s)e^(-lambda233280s)
200 = e^(-lambda233280s)
ln200 = -lambda233280s
lambda = 2.2712x10^-5 s^-1

R = lambdaN
7.4x10^12 decays/s = (2.2712x10^-5 s^-1)N
N = 3.2582x10^17 nuclei

Please let me know if I have done the correct steps up to this point. I do not know what to do next. Thanks.

Last edited: Dec 13, 2006
4. Dec 14, 2006

### daveb

The first equation you list is the equation for activity of a sample given some initial activity, and is commonly called the decay equation. To find the decay constant (lambda), you use the second equation since you already know the half life. The equation you don't have listed is that Activity of any given sample is the number of atoms in the sample times the decay constant. Since you now have the number of atoms, you can use Avogadro's number and the gram atomic weight to determine the mass.
Your lambda in this case is incorrect because lambda = ln2/half-life = .693/2.70 days = 0.000002971 per second. The way you set up your equation, you wrote that 200 curies = 1 curie*e(-lambda*half-life).

Last edited: Dec 14, 2006
5. Dec 14, 2006

### OlderDan

You are correct that your R is my A, and my constant is your λ. The 200 Ci activity has nothing to do with the calculation of λ from the given half life. daveb is correct on that point. But you do have the equation he says is missing in your quote above.

6. Dec 14, 2006

### FlipStyle1308

Okay, so here's my new work that I just did:

λ = ln2/T1/2 = 0.693/233280s = 2.9707x10^-6 decays/s
R = λN
200Ci x 3.7x10^10decays/s/Ci = (2.9707x10^-6decays/s)N
N = 2.491x10^18
(2.491x10^18)(6.022x10^23molecules/mol) = 1.5x10^42

I did not quite understand what you meant when I incorporate the gram atomic weight into this because it is not given. (The atomic mass in the book's appendix only lists that for 197_79_Au instead of 198_79_Au, which I need.

7. Dec 14, 2006

### OlderDan

http://chemlab.pc.maricopa.edu/periodic/Au.html

197.96822

8. Dec 14, 2006

### FlipStyle1308

Thanks. But I still don't know how to plug this into whatever equation I'm supposed to be using. Or is it...

(2.491x10^18)(6.022x10^23molecules/mol)(197.96822) = 2.9697 x 10^44?

Last edited: Dec 14, 2006
9. Dec 14, 2006

### OlderDan

Your N should be the number of molecules, so you should be dividing by Avegadro's number to find the number of moles. The 197.96822 is molar mass, or grams per mole, so I believe you are looking for

(2.491x10^18molecules)(197.96822gm/mol)/(6.022x10^23molecules/mol)

10. Dec 14, 2006

### FlipStyle1308

So I should get an answer of 0.8189 mg?

11. Dec 14, 2006

### OlderDan

Again you want somebody to multiply for you?