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.
Ro = kNo
k = 0.693/t1/2 = 0.693/7.36E15 = about 9.4E-17
Ro = 9.4E-17 x 6.62E23 = ?dpm
Find the number of disintegrations per minute emitted by 1.1 mol of 232-Th in 1 min. The half-life of 232-Th 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 half-lives = ending amount (sometimes remaining rather than ending is used)
(1/2)number of half-lives is the decimal fraction which remains (1.000 is the original starting amount, 0.500 at the end of one half-life, 0.250 at the end of two, 0.125 at the end of three, etc.)
number of half-lives that have occurred can be expressed as (total time elasped ÷ length of half-life)
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