Posted by **sarah** on Saturday, April 27, 2013 at 6:29pm.

The half-life of is 30.0 days. What fraction of a sample of this isotope will remain after 16.2 days?

- chem -
**jimbo**, Saturday, April 27, 2013 at 7:28pm
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%

- chem -
**DrBob222**, Saturday, April 27, 2013 at 8:15pm
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/t_{1/2}. 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.

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