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Chemistry: Rates of Disintegration Reactions

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  • Type: Video Tutorial
  • Length: 14:49
  • Media: Video/mp4
  • Use: Watch Online & Download
  • Access Period: Unrestricted
  • Download: MP4 (iPod compatible)
  • Size: 159 MB
  • Posted: 07/14/2009

This lesson is part of the following series:

Chemistry: Full Course (303 lessons, $198.00)
Chemistry: Nuclear Chemistry (8 lessons, $12.87)
Chemistry: Rates of Disintegration (2 lessons, $3.96)

This lesson was selected from a broader, comprehensive course, Chemistry, taught by Professor Harman, Professor Yee, and Professor Sammakia. This course and others are available from Thinkwell, Inc. The full course can be found at http://www.thinkwell.com/student/product/chemistry. The full course covers atoms, molecules and ions, stoichiometry, reactions in aqueous solutions, gases, thermochemistry, Modern Atomic Theory, electron configurations, periodicity, chemical bonding, molecular geometry, bonding theory, oxidation-reduction reactions, condensed phases, solution properties, kinetics, acids and bases, organic reactions, thermodynamics, nuclear chemistry, metals, nonmetals, biochemistry, organic chemistry, and more.

Dean Harman is a professor of chemistry at the University of Virginia, where he has been honored with several teaching awards. He heads Harman Research Group, which specializes in the novel organic transformations made possible by electron-rich metal centers such as Os(II), RE(I), AND W(0). He holds a Ph.D. from Stanford University.

Gordon Yee is an associate professor of chemistry at Virginia Tech in Blacksburg, VA. He received his Ph.D. from Stanford University and completed postdoctoral work at DuPont. A widely published author, Professor Yee studies molecule-based magnetism.

Tarek Sammakia is a Professor of Chemistry at the University of Colorado at Boulder where he teaches organic chemistry to undergraduate and graduate students. He received his Ph.D. from Yale University and carried out postdoctoral research at Harvard University. He has received several national awards for his work in synthetic and mechanistic organic chemistry.

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Thinkwell
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We've talked about the energetic of nuclear reactions and showed that what drives nuclear reactions is increasing binding energy per nuclei. So, in other words, the nucleus becomes more stable. Just like in chemical reaction, products are more stable than reactants if you have an exothermic reaction. In a nuclear reaction, products are more stable because the binding energy per nucleon increases. And that was drives not only fusion and fission, but it also drives alpha decay and beta decay and electron capture and positron emission. That's the idea that the nucleus becomes more stable after it decays. Well that says absolutely nothing about the rate at which the reaction occurs. Remember, we've talked about thermodynamics and kinetics and how thermodynamics and kinetics are largely independent of each other, that even if you have a lot of driving force for a reaction, it could be as slow as a glacier.
So what about the kinetics for nuclear reactions? Well, all the nuclear reactions that I'm aware of are first-order. That's to say that the rate at which daughter is created only depends on how much parent you have. And we expressed a unique quantity that describes how fast first-order reactions as the half-life - the time it takes for half the sample to radioactively decay - or in the chemical reaction, the time it takes for half of the reactant to go away.
So we have strontium-90 going to yttrium-90 plus an electron - this is an example of a beta decay where the half-life is 29 years. That says if we start with 10 grams of strontium, after 29 years, we'll only have 5 grams. After another 29 years, we'll only have 2 grams and so on and so on and so on. Strontium-90 turns out to be sort of bad for you when it's in the environment because its chemical properties, being an alkaline earth, are very much like those of calcium, and so it can get incorporated into cow's milk, for instance, then you take the cow's the milk and it incorporates strontium-90 into your bone. That's a route to bone cancer and things like that.
This idea of a half-life - and it's a measured quantity, so someone has just gone through and measured the half-life of all these different radioactive nuclei - explains why there is uranium-238 and uranium-235 naturally occurring, even though they're both radioactive. They naturally occur, but all of the heavier nuclei don't exist in nature and the reason is, it turns out, you could not predict this a priori. So, in other words, you couldn't know that this would be true, but you can measure it. The half-life of uranium-238 and the half-life of uranium-235 are both very long. 238, for instance, is 4.5 billion years. And this is on the order of the understood, at least, age of the planet. So when the planet was formed, we had some amount of and we had some amount of and we had some amount of plutonium-239 and all the other elements, at least in principle, but they decay. And they decay such that after, say for strontium-90, if we had only strontium-90 that we started with when the planet was formed, then after 29 years, only half of it would be gone. Well, after 4 billion years, or however old the planet is, all that strontium-90 is going to be gone if you don't make it anymore. Well the understood idea is that, in nuclear synthesis, to make these heavy elements, that they were formed a long time ago. And however much we started with on the planet originally, that was how much there was. And so uranium-238 and 235 - there's still a little bit - in fact, there's a fair bit of uranium-238 because it's half-life is only on the order of the age of the planet. Whereas, for instance, plutonium-239, after 24,000 years, half of it is gone. After another 24,000 years, half more of it is gone. So there's just not any leftover from when the planet coalesced.
Well, this also explains why we have a nuclear waste problem, because we've made plutonium-239. So we use this as a nuclear fuel - and we're going to talk about that later on - and we use it to make bombs. And after the usefulness of the fuel rod is gone, we still have to deal with the waste that's left over. It's not good enough to make electricity, but we have to deal with it. And not only do we have to deal with it, we have to deal with it for 24,000 years. But really, after 24,000 years, only half of it is gone. And realistically, we have to wait about 10 half-lives, or 240,000 years, before its radioactivity has decayed to a reasonably safe level. After ten half-lives, it's down to about of what it was originally. Right now, our best solution to dealing with nuclear waste is just to bury it. You can't really do a chemical transformation and make it any safer, because plutonium metal, plutonium oxide, plutonium whatever is still going to behave like plutonium-239, because it's a nuclear reaction not a chemical reaction. Heating it up, for instance, doesn't make this reaction go faster, because it's so ridiculously slow.
So, let's go back to the first-order kinetics review that I was talking to you about. And we've got, for first-order reaction, that the concentration of the reactant at any time t divided by the concentration at time zero is equal to minus k x t, where k is the first-order rate constant. The rate expression is given by rate is equal to k times the concentration at any time t. So the rate of the reaction depends on first-order rate constant times the concentration of the reactant. And we can take this expression and re-express it by taking the exponential of both sides and then rearranging. And we get that the concentration of reactant at any time t is equal to the concentration originally times the exponential of minus k x t. So in other words, the concentration of A has an exponential decay. And you'll recall that we derived an expression for the half-life by setting the concentration of A at some time t[1/2] to be of what it started out. Go back and review that if you don't remember it. But the point is that we have that the half-life is equal to the natural log of 2 divided by the first-order rate constant. And the natural log of 2 has a value of 0.693. So this is an expression that relates half-life to the first-order rate constant.
Okay, so the analogous equations for radioactive decay, instead of talking about concentrations, we're just going to talk about numbers of nuclei. So here's the number of nuclei initially in the sample, and it decays with the first-order rate constant, k, such that the number of radioactive nuclei at some time t is given by this equation. We have the same expression for the half-life. And we can define a quantity called the activity. And the activity is equal to first-order rate constant times the number of nuclei we have that are radioactive. And this is exactly parallel to rate is equal to k times the concentration of A, except now we have a rate of nuclear decay is equal to the first-order rate constant times the number of parents. N is the number of parents. Now, since we have this relation in purple and this relationship in green, we can relate those two into an equation in pink here that says that the activity at any time t is equal to the activity originally times the exponential of minus k x t, where k is the first-order rate constant.
Now, what is activity? Well, it's measured in disintegrations per unit time. And the SI unit is the becquerel and 1 becquerel is defined to be 1 disintegration per second. Now an older unit that's not SI, but it's still very commonly used, is the curie - abbreviated Ci - and it's equal to 3.7 x 10^10 disintegrations per second. That's 37 billion disintegrations per second. You can't help but wonder who sat around counting 37 billion disintegrations per second. Obviously, this is a huge number and so when you see curies, most often you see millicuries, which are of a curie or microcurie, which would be of a curie. Well, how do we measure these? How do we measure disintegrations? And one of the answers is that we use something called a Geiger counter. And if you take a look at the graphic, you'll see a schematic of the Geiger counter. And it takes advantage of the fact that radioactive decay gives rise to particles - alpha particle, beta particle - that are ionizing. In other words, when they strike a neutral atom, they have so much kinetic energy that they will wipe one or more electrons off of that neutral atom.
So, inside the probe of a Geiger counter - and let me turn this guy on - and this is the probe, there is an anode and a cathode and those two create a voltage. And in the absence of anything in the gas - so there's argon gas on the inside as well - and in the absence of any ions in the gas phase, no current is passed and so the thing is absolutely quiet. But every now and then, something is going to strike an argon atom and ionize it. And when that argon atom is ionized, it's going to be very quickly attracted, because it is positively charged. But it will be very quickly attracted to the negative electrode, completing the circuit, and what you'll hear is a click. Now, if you listen carefully, you'll hear an occasional click. I'll be quiet for a second. And what that is, that's the background radiation. Remember, we're being inundated by radiation all the time - from outer space - but also, it's just naturally occurring. We have potassium-40 in our bodies, and that's radioactive. We have boron in basements, and that's radioactive. So there is some level of background radiation. But there's a source of radiation in most of your houses - hopefully in all of your houses - and that is a smoke detector. A smoke detector uses americium-241 and the way it uses it is that the americium alpha decays, and so it spits out this alpha particle, and the alpha particle goes to complete a circuit. So, so long as the americium is decaying, there is some average voltage in that current in the circuit and everything is hunky dory. But what happens is if smoke comes along, it starts to get in the way of these alpha particles that are being emitted from the source. And so what happens is the current drops and when the current drops, the smoke detector makes a lot of noise. That's why when the battery dies - the battery is the thing that drives the circuit - and when the battery dies, typically things go bad. The smoke alarm starts to make noise. And the reason is that it's signaling that the current is dropping.
So let's listen to how many decays these things make. And it's not much higher than the background, so I warn you, obviously, it doesn't make sense to have a really hot source inside your house. Okay, so that's a little bit above background. Now, let's do a calculation.
Actually, let me say one last relationship. If we rearrange this equation and we combine it with this equation - so green and purple - we have N is equal to A, the activity, divided by k, the rate constant, and we have an expression for the rate constant in terms of the half-life. So the number of radioactive nuclei in our sample is equal to the activity times the half-life divided by the natural log of 2. And so we can use that equation. If we know activity and we know the half-life, we can know how many nuclei we have in our sample that are radioactive.
So a sample of americium-245 with a half-life of 7.37 x 10^3 years, or a little over 7000 years, we look these numbers up in tables like the CRC. And it has an activity of 1.00 becquerels. So our sample was clicking at about 1 per second, roughly, and so that's what we use as our activity. Calculate the number of americium-245 nuclei initially and the number after 100 years. Well, how do we do that?
First of all, we know what the half-life is in years. Let's convert that to seconds. How we do that is we multiply by the number of seconds of in a minute, the number of minutes in an hour, the number of hours in a day and the number of days in a year. And so we get 2.32 x 10^11 seconds. And then we use the equation that I wrote up that he number of radioactive nuclei is equal to the activity times the half-life divided by the natural log of 2. And so we plug in for the activity, which was 1 disintegration per second - to get the units right, it would be 1 nuclei disintegrating per second. And then, the half-life divided by the natural log of 2, and we get 3.35 x 10^11 nuclei. So in our sample, that's how many americium-245's we have.
And now we can ask how many are going to be left after 100 years, and to do that, we use this equation, which is something I've also given you, that the number after 100 years is equal to the number we have today times e^-kt where t is 100 years and k is given by the rate constant for the problem. And the rate constant for the problem is the natural log of 2 divided by the half-life. So .693 divided by 7.37 x 10^3, which is equal to 9.4 x 10^-5 inverse years. And we're keeping it in years now because we want to know how much is going to be around after 100 years. So we got to get the units to cancel out. And then, N is equal to the number that we have now, which I calculated before, times . That's the rate constant in inverse years times 100 years, which is how long we were going to go. And that gives you 3.32. So basically, 99 percent of our sample is still there. And now, let's go ahead and calculate the activity, just as an additional exercise. The activity after 100 years is equal to the activity today times the exponential, minus kt, just like the equation for N. And so if our activity is 1.00 becquerel today, it's going to be .991 becquerel in 100 years.
So, I've showed you how you can use the fact that it's a first-order rate of decay. And if we know the half-life, we can calculate k. And using that and the activity, which is measurable, we can find out how many atoms we have that are radioactive and we can make predictions about how radioactive something is going to be in the future as well, again, using these algebraic expressions.^
Nuclear Chemistry
Stability of Atomic Nuclei and Rates of Distintegration
Rates of Disintegreation Reactions Page [3 of 3]

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