Posted by Nancy on .
Find the number of disintegrations per minute emitted by 1.1 mol of 232Th in 1 min. The halflife of 232Th is 1.4 x 1010 year
I was able to do this with the following technique:
1.1 x 6.022 x10^23 =6.62x10^23
1.4x10^10=7.358x10^15 minutes
I think then it is 6.62x10^23 x 7.358x10^15
But I want to learn how to do this problem using the formulas below instead of the .693/half life.
There are the following formulas:
starting amount) x (1/2)number of halflives = ending amount (sometimes remaining rather than ending is used)
(1/2)number of halflives is the decimal fraction which remains (1.000 is the original starting amount, 0.500 at the end of one halflife, 0.250 at the end of two, 0.125 at the end of three, etc.)
number of halflives that have occurred can be expressed as (total time elasped ÷ length of halflife)

chemistry 
DrBob222,
I think what you want applies only to what remains or what decays and not to activity.
If we want to know what remains after 2 half lives it is (1/2^n) = 1/4 or after 3 half lives is (1/2^3) = 1/8 etc.
What you have here is activity at t=0.
R^{o} = kN^{o}
k = 0.693/t_{1/2} = 0.693/7.36E15 = about 9.4E17
Ro = 9.4E17 x 6.62E23 = ?dpm