Kinetics of radioactive decay | Kinetics | AP Chemistry | Khan Academy

Strontium-90 is a radioactive isotope that undergoes beta decay. Because radioactive decay is a first-order process, radioactive isotopes have constant half-lives. Half-life is symbolized by t1/2, and it's the time required for one half of a sample of a particular radioactive isotope to decay. For example, the half-life of strontium-90 is equal to 28.8 years.

Let's say we start with 10 grams of our strontium-90 radioactive isotope. On the y-axis, we're going to put the mass of our strontium isotope in grams, and on the x-axis, we're going to have time. When time is equal to zero, we have 10 grams of our isotope. Since the half-life of strontium-90 is 28.8 years, if we wait 28.8 years, we'll go from 10 grams to 5 grams. So, the next point on our graph would be at 5 grams, and this time should be 28.8 years.

If we wait another 28.8 years, we're going to go from 5 grams to half of that, which would be 2.5 grams. So, the next point would be here at 2.5. If we wait another 28.8 years, we go from 2.5 grams down to 1.25, so approximately here on our graph. This graph shows exponential decay.

Let's say we were asked to find out how much of our radioactive isotope of strontium is left after 115.2 years. What we would do is take 115.2 years and divide that by the half-life of 28.8 years. By doing that, we realize that 115.2 is really just four half-lives. One approach to this problem will be starting with our 10 grams, and we think about one half-life taking us to 5 grams, another half-life taking us to 2.5, another half-life taking us to 1.25, and then finally one more half-life that takes us to 0.625 grams.

So, just doing the problem this way, we can see that's one, two, three, four half-lives. Our final answer is that 0.625 grams remain after 115.2 years. Another approach to do the same problem would be to start with our 10 grams, and we multiply that by one-half to get the amount remaining after one half-life. We could do that three more times to get the amount that remains after four half-lives. We could have also written this as 10 times one-half to the fourth power since we needed to wait four half-lives.

All these approaches will get you the answer of 0.625 grams of our radioactive isotope remaining after 115.2 years. For a chemical reaction with reactant A that's first order, the rate law says that the rate of reaction is equal to the rate constant k times the concentration of A to the first power. Since radioactive decay is a first-order process, we can write that the rate of decay is equal to the rate constant k times N to the first power, where N is the number of radioactive nuclei in a sample.

Since radioactive decay is a first-order process, we can also use this equation for the rate constant, which comes from first-order kinetics. It says that the rate constant k is equal to 0.693 divided by the half-life. For example, if we wanted to find the rate constant for the radioactive decay of strontium-90, the rate constant would be equal to 0.693 divided by the half-life of strontium-90, which we saw was 28.8 years.

So, when we do that math, we find that k is equal to 0.0241 1 over years. Another equation from first-order kinetics is the integrated rate law for a first-order reaction. The integrated rate law says that the natural log of the concentration of reactant A at some time t is equal to negative kt plus the natural log of the initial concentration of reactant A.

Since we're using N, which is the number of radioactive nuclei in our sample instead of the concentration of A, we can write the integrated rate law for a first-order radioactive decay process. This says that the natural log of the number of radioactive nuclei at some time t is equal to negative kt plus the natural log of the initial number of radioactive nuclei.

Let's say we start with 1.000 grams of our strontium-90 radioactive isotope, and our goal is to find out how much remains after two years. So, we're going to wait two years and find out how much of our radioactive isotope remains. We're going to use our equation for the integrated rate law.

Let's go ahead and plug in what we know. We already know what k is; we found that in our earlier problem. So, we can write this as equal to negative k, which is 0.0241. Next, we also know the time period we're interested in, so we know what t is. t is two years, so let's go ahead and write 2.00 in here for the time.

Next, we're going to add the natural log of the initial number of radioactive nuclei in our sample. While we don't have the initial number, we do have the mass, and since the mass is proportional to the number of radioactive nuclei, it's okay to go ahead and plug that into our equation. We're going to plug in the natural log of 1, and all of this is equal to the natural log of the initial number of radioactive nuclei at some time t.

So, we have the natural log of that. The natural log of 1 is zero. Now we have the natural log of N is equal to… and when we do this math, we're going to get negative 0.0482. Next, we have to get rid of the natural log, and we can do that by taking e to both sides. So if we take e to both sides, the natural log cancels out, and we get that N is equal to 0.953 grams.

So that's how much of our radioactive isotope remains after two years.