posted by sarah .
The half-life of is 30.0 days. What fraction of a sample of this isotope will remain after 16.2 days?
after 30 days 50% of the sample would have decayed. 50% / 30 days gives you a daily decay rate of 1.6666667. multiply this by 16.2 to give 27. 27 is the percentage of the sample that has decayed after 16.2 days so 100-27 = 73%
I don't believe that 1.6667 is a constant. For example, at the end of another 30 days it shows 50% and that + te 50 already gone is now zero remaining when in fact there is 25% remaining after 60 days.
k = 0.693/t1/2. Solve for k and substitute into the equation below.
ln(No/N) = kt
No = 100%
N = % remaining
k from above
t = 16.5 days.
Solve for N.