This example problem demonstrates how to use an isotope's half life to determine the amount of the isotope present after a period of time.

### Half Life Problem

^{228}Ac has a half life of 6.13 hours. How much of a 5.0 mg sample would remain after one day?

### How To Set Up and Solve a Half Life Problem

Remember that the half-life of an isotope is the amount of time required for one-half of the isotope (the parent isotope) to decay into one or more products (daughter isotope).

In order to work this type of problem, you need to know the decay rate of the isotope (either given to you or else you need to look it up) and the initial amount of the sample.

The first step is to determine the number of half lives that have elapsed.

number of half lives = 1 half life/6.13 hours x 1 day x 24 hours/day

number of half lives = 3.9 half lives

For each half life, the total amount of the isotope is reduced by half.

Amount remaining = Original amount x 1/2^{(number of half lives)}

Amount remaining = 5.0 mg x 2^{-(3.9)}

Amount remaining = 5.0 mg x (.067)

Amount remaining = 0.33 mg**Answer:**

After 1 day, 0.33 mg of a 5.0 mg sample of ^{228}Ac will remain.

### Working Other Half Life Problems

Another common question is how much of a sample remains after a set amount of time. The easiest way to set up this problem is to assume you have a 100 gram sample. That way, you can set up the problem using a percentage.

If you start with a 100 gram sample and have 60 grams remaining, for example, then 60% remains or 40% has undergone decay.

When performing problems, pay close attention to the units of time for half-life, which might be in years, days, hours, minutes, seconds, or tiny fractions of seconds. It does not matter what these units are, so long as you convert them to the desired unit at the end.

Remember there are 60 seconds in a minute, 60 minutes in an hour, and 24 hours in a day. It's a common beginner mistake to forget time isn't usually given in base 10 values! For example, 30 seconds is 0.5 minutes, not 0.3 minutes.