[clockworks204]

Actinium (²²⁷Ac) has a half life of 22 years.  One gram of actinium contains 2.65x10²¹ atoms.  What is the activity of one gram of actinium? What is the activity is curies? (one curie is 3.7x10¹⁰ disintegrations per second)


I know that the formula for unstable atoms left over after t time is N=N_oe^{-0.693t/t_1/2}
I know that t_1/2= 22 yrs, N_o= 2.65x10²¹ atoms.  I'm not sure where to go from here.

[MrTeo]

The formula you've written comes from the solution of this differential equation:



Here as you can see the first term is the activity of the amount of compound given (disintegrations per second) so you only need to find out the decay constant (if you're not sure about that continue reading) and then you'll easily find the activity in Curie as the problem asks you.

Bonus:
Here is the known formula and how you can get it from the differential equation I gave you:



Knowing that at t=0 we have N₀ atoms gives us the final formula:



Finally the relation with half-life:

[clockworks204]

Thank you for your reply.  I've been studying what you've posted for a bit, and have tried applying it.   
k=.693/22 years = .0315 for the decay constant, and use your first equation to multiply .0315 by 2.65x10²¹ to get 8.3x10¹⁹.

According to the textbook, the answer for this should be 2.6x10¹².  The activity in curies is supposed to be 71 curies.

I've tried this problem for a while now, and still can't come up with the answers from the textbook

[MrTeo]

Just convert 22yr to seconds

[clockworks204]

Got it!  You've been a great *delete me*