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### Course: Chemistry archive > Unit 10

Lesson 2: Relationship between reaction concentrations and time- First-order reactions
- First-order reaction (with calculus)
- Plotting data for a first-order reaction
- Half-life of a first-order reaction
- Worked example: Using the first-order integrated rate law and half-life equations
- Second-order reactions
- Second-order reaction (with calculus)
- Half-life of a second-order reaction
- Zero-order reactions
- Zero-order reaction (with calculus)
- Kinetics of radioactive decay
- 2015 AP Chemistry free response 5

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# Kinetics of radioactive decay

The nuclei of radioactive elements decay according to first-order kinetics. As a result, the half-life equation and integrated rate law for radioactive decay processes can be derived from the rate laws for first-order reactions. The resulting equations can be used to find the rate constant

*k*for a decay process and determine the amount of radioactive isotope remaining after a certain time period. Created by Jay.## Want to join the conversation?

- Why is radioactive decay a first-order reaction?(4 votes)
- Radioactive decay only depends on the concentration of a single reactant; the radioactive atom. It doesn't need to react with another reactant for decay to happen. If our reaction rate depends on the concentration of only a single reactant then that's first-order.

Hope that helps.(10 votes)

- why would it be ok to put the gram amount into the equation in place of radioactive nuclei? even if its proportional, you would get a really different number(3 votes)
- The amount of radioactive material remaining after a certain time, [A]t, can be any unit which describes the amount of stuff there is. So we could use number of atoms, moles, grams, or kilograms for example. If we use grams as our initial material, [A]0, then the final amount will also be in grams. Likewise, if we start with number of atoms, we’ll calculate the final in number of atoms as well. Of the two though, number of atoms and grams, grams are easier to use since we can measure a radioactive sample in grams much easier than we can its number of radioactive atoms.

Additionally, we can convert from grams to number of atoms if we really need to.

Hope that helps.(3 votes)

- When I am finding the radioactive decay of a given time, can I use the integrated rate law when I have identified the rate constant, k. I identify the k from the half-life formula for first order reaction. The reason why I am asking is that when I got an answer form integrated rate law, it is slightly inaccurate as I got 6.80, when the correct answer was 6.79. I just want to know why I cannot use just plug in the k into integrated rate law and find the answer. Also I do not quite understand the way that khan academy wants me to solve these types of problems. Thanks so much!!(1 vote)

## Video transcript

- [Instructor] 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 1/2 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. And on the y-axis, we're gonna put the mass of
our Strontium isotope in grams. And on the x-axis, we're
going to have time. So 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 five grams. So the next point on our
graph will be at five grams, and this time should be 28.8 years. If we wait another 28.8 years, we're gonna go from five
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'd go from 2.5 grams down to 1.25, so approximately here on our graph. So this graph shows exponential decay. Let's say we were asked to find out how much of our radioactive
isotope Strontium is left after 115.2 years. So what we would do is take 115.2 years and divide that by the
half-life of 28.8 years. And by doing that, we realize that 115.2 is
really just four half-lives. So one approach with this problem would be starting with our 10 grams. And we think about one half-life
taking us to five 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. So our final answer is 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 1/2 to get the amount remaining
after one half-life. And we could do that three more times to get the amount that
remains after four half-lives. We could have also have written this as 10 times 1/2 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, which 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 one over years. Another equation from first-order kinetics is the integrated rate law
for a first-order reaction. And 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 our first-order
radioactive decay process, which says 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. So 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 gonna wait two years and find out how much of our
radioactive isotope remains. So 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 is
equal to negative times 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 gonna add the natural log of the initial number of
radioactive nuclei in our sample. And 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. So we're gonna plug in
the natural log of one. 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 one is zero, so now we have the natural log of N is equal to, and when we do this math, we're
gonna 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.