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Radioactive Decay | 40 mins | 0 completed | Learn |

Band of Stability | 17 mins | 0 completed | Learn |

Magic Numbers | 10 mins | 0 completed | Learn |

First Order Half Life | 19 mins | 0 completed | Learn |

Radioactive reactions follow principles dealing with Chemical Kinetics.

Concept #1: Understanding Positron Emission.

**Transcript**

Hey guys! In this new video, we're going to take a look at radioactive decay rates. We’re going to say when it comes to radioactive decay, it’s going to follow a first-order mechanistic step. We’re going to say here it's a first-order process whose rate is proportional to the number of radioactive nuclei N, where k is a first-order rate constant called the decay constant. This kind of follows what we learned earlier when it comes to kinetics. If you’ve seen my earlier videos on chemical kinetics, you should be very familiar on how exactly we’re going to solve these types of questions because when it came to chemical kinetics, we did the Integrated Rate Laws. We’re going to say when it comes to radioactive decay, we're going to basically follow those Integrated Rate Laws just a little bit differently.

When it comes to kinetics, we’re going to say that Rate Law equals our rate constant k times the concentration of our reactant to some power. That power we call a reaction order. Since we’re dealing with radioactive systems now, it's basically this equation just changed a little bit. Here we still have rate but instead of saying Rate Law, we’re saying Decay Rate. Here, this is our rate constant k but now it’s our decay constant k. Here, this was the concentration of our reactant, but now it's the concentration of our radioactive nuclei. Similar but yet different.

Now, we’re going to say again radioactive decay follows a first-order process like I said earlier. Here we’re just going to reuse the first-order integrated rate law from the chapter on chemical kinetics. Remember, kinetics just looks at rate or speed of reactions. We're looking at how quickly do my reactants break down to give me a product. But since we're talking about radioactive systems now, we’re saying how quickly will my radioactive compound breakdown. That’s just the fundamental difference.

In regular kinetics, we’re looking at how fast the reactant basically becomes product. Here we're looking at how fast is my radioactive compound just decay away. Here, when we dealt with chemical kinetics, we said that the integrated rate law for the first-order process was ln of my final concentration of my reactant equals negative rate constant times time plus ln of the initial concentration of my reactant. Again, this was your final concentration of your reactant. This right here was your rate constant. Here this was your time and here this was the initial concentration of your reactant. Since we're going to need some room guys to finish writing all that notes that we need, I'm going to remove myself from the image.

Remember, this was in chemical kinetics, the first-order integrated rate law. But now we're dealing with radioactive systems, so the equation is going to change a little bit. Here we’re going to have ln Nt. Here. Nt stands for what? We're going to say here Nt stands for the final concentration of our radioactive nuclei. Just think of it as your radioactive reactant. Radioactive nuclei is equivalent in a sense to your radioactive reactant. That equals negative kt plus ln of No. You can see that they're very similar. All we’re really doing is substituting in N and replacing the variable A. That’s the only real difference here.

Here instead of this being our rate constant, it's going to be now our decay constant. Here this is still time. Here, this would just represent your initial concentration of your radioactive nuclei. But they're very similar equations. If you know how to use the one from the chemical kinetics chapter, then you’ll have no problem adopting it to these radioactive systems because it's basically the same type of process.

Radioactive decay follows a 1^{st} order mechanism and so has a similar integrated rate law equation from Chemical Kinetics.

Example #1: A sample of radon-222 has an initial α particle activity (A_{0}) of 8.5 x10^{4} dps (disintegrations per second). After 7.3 days, its activity (A) is 3.7 x 10^{4} dps. What is the half-life of radon-222?

**Transcript**

Here, since we're going to continue to need more space guys, I’m going to stay out of the image a little bit longer. Here we're going to say: A sample of radon-222 has an initial alpha particle activity. Just think of that as your initial concentration of 8.5 times 10 to the 4, that should be a superscript, dps which means disintegrations per second. Basically it's breaking down by that much per second. After 7.3 days, its activity A is 3.7 times 10 to the 4 disintegrations per second. What is the half-life of radon-222?

Again, we're dealing with first-order kinetics here. But now it's just first-order radioactive kinetics. Here it's going to be ln Nt equals negative kt plus ln No. Here they're asking is for half-life. Remember for first-order, half-life was half-life equals ln 2 divided by k. Here we’re going to say that ln 2 is the same thing as 0.693 when you punch it into your calculator. They're asking me to find half-life so what I need to find is k. Remember, we're dealing with first-order process is because radioactive decay occurs by a first-order process so it has the same exact half-life whether you're dealing with radioactive systems or not. All we got to do now is just plug in what we know.

I tell you that initially this is what you have, so this is your initial concentration. Then we're going to say here that this is your activity after a given amount of time, 7.3 days to be exact. That represents your final concentration equals, we don't know what k is. That's what we're looking for, so negative k. Time is 7.3 days. Remember, because your time is in days, that means your k will be in days’ inverse. They have to agree in terms of time. If k is in minutes’ inverse for example, time has to be in minutes. They have to agree.

All we got to do now is separate just the k by itself. What we're going to do next is we're going to subtract ln 8.5 x 10 to the 4 from both sides here. This gets taken out. Over here, it’s going to give me negative 0.831733. That equals negative k times 7.3 days’ inverse. Next, we want to just isolate k by itself, so we're going to divide out negative 7.3 days from both sides here. The negative gets cancelled out and the 7.3 days gets cancelled out now. Because your days is on the bottom, when we find k it will be in days’ inverse. The negative and the negative cancel out so I have 0.114 days’ inverse. Those days’ inverse, I just plug in over here 0.114 days’ inverse and because days is on the bottom here, when I find half-life, it's just going to be in days because I'm bringing it up back up top. Here it will give you 6.08 days.

We might be talking about radioactivity now, but fundamentally it's still the same thing when it came to chemical kinetics. We're dealing with a first-order process, so we're using it a sense the first-order integrated rate law with the half-life for a first-order reaction. If you can just see it as that, it's the same exact things that we've done earlier before in the earlier chapters. Remember that and just apply it so that you can get through problems like this. Here instead of talking about molarity of our reactant, we're talking about activities, we're talking about disintegrations per second. It's the same thing. Just apply what you learned earlier to what we're doing now.

Example #2: Gallium citrate, containing the radioactive isotope gallium – 67, is used medically as a tumor seeking agent. It has a half – life of 78.2 hours. How long will it take for a sample of gallium citrate to decay to 20.0% of its original activity?

Example #3: What percentage of carbon – 14 ( t_{1/2} = 5715 years) remains in a sample estimated to be 16,230 years old?

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