Radioactive Half-Life - Video Tutorials & Practice Problems

Intro to Radioactive Half-Life

Here, we can say that radioactive half lives, which is tea, sub half, it's just the amount of time required for half of a radioisotope to decay. Now, a radioisotope or nuclide is just an isotope that has an unstable nucleus and emits radiation as it decays alpha decay, beta decay, electron capture positron emission, et cetera based on this. If we take a look here, the example says, what is the half life of the radio isotope that shows the following data of remaining percentages versus time. So here we have our percentage remaining and here we have a number of days we start out with 100% of our starting material on day zero. Remember half life looks at how much time it takes us to lose half. All right. So 100 half of that would be 50 half of that would be 25%. Another half of that would be 12.5%. What are the number of days that separate each one of these percentages? Well, day zero, it takes three days for us to lose half looks like it takes another three days to lose half and it looks like it look, took another three days to lose half. So based on this chart here, it looks like the half life is equal to three days for this particular radioisotope.

Method 1 of Radioactive Half-Life

Now we're gonna look at different methods that involve half life in some way. In method one, we're dealing with the direct calculation of half life or rate constant. Now, in method one, you use the radioactive half life equation when dealing with only the half life and the decay constant, which is K. Now here, the formula for the radioactive, half life is half life equals Ln two over KLN two is a constant, which is equal if you plugged into your calculator to approximately 0.693 K, here is our decay constant, which is in times inverse, which shouldn't be mistaken with time, which is t here. When it comes to a plot of half life versus time, we would say that half life does not depend on the initial concentration because if we look, we don't see this variable within the formula and it is constant, do the whole reaction. Remember the example that we saw earlier every three days, we lose half of our starting material. It was every three days. It didn't fluctuate where it's 2.5 here and 1.8 here or 7.2 here. It was always three days because of that. Well, if we were to plot this, we'd say that the half life remains flat, it's constant. So no matter how much time passes, my half life stays the same. So it just be this flat straight line. So when it comes to the radioactive half-life equation, just remember, this is the equation we utilize. It forms a connection between half life and our decay. Constant K, half life is a constant idea because we're following first order rate law rules.

Radioactive Half-Life Example

Here, we're told if the decay constant of plutonium 244 is 8.66 times 10 to the negative nine years at 25 °C. What is its half life? All right. So they're asking for half life and all they're giving us is the decay constant. This is just simply utilizing the radioactive half life equation where half life equals Ln two over K plug in the value that we just got was 8.66 times 10 to the negative nine years. Inverse should be inverse. That will give me 8.00 times 10 to the seven years. So this would be our final answer. This would be equal to the half life of plutonium 244. It would take this amount of time for us to lose half of our starting material.

Radioactive Half-Life Example

Now method two deals with radioactive nuclei concentrations. In method two, we utilize both the radioactive half-life equation and the radioactive integrated rate law. So if we look here's our half-life equation and here's our integrated rate law. In this case, the questions will involve the half life time, your initial concentration or your final concentration for your radioactive nuclei concentrations. If we take a look at this example, it says a sample of radon 222 has an initial alpha particle activity. So here a sub knot now this can just be a stand in for our variable of N it's the same idea here it has as its concentration disintegrations per second. So we have 8.5 times 10 of the four disintegrations per second. After 7.3 days, its activity now becomes 3.7 times 10 of the four disintegrations per second. What is the half life of radon? 222? All right. So half life equals Ln two over K. So for us to know what the half life is, we need to determine what our decay constant will be. From this question, we have what we have this larger concentration which is our initial, we have the smaller one after X number of days, which is our final. And we have time looking at the integrated rate law, we have this variable, this variable and this variable, knowing those three variables, we can solve four K. Once you solve for K, you can plug it into the half-life equation and we'll know half life. So here we're gonna say all N of final amount, final concentration equals negative KT plus all line of initial concentration. So here we'd say all of 3.7 times 10 to the four equals negative K which we don't know, we have 7.3 days plus IGN of initial. We're gonna subtract allen of our initial from both sides. When we do that, here, we're gonna get initially negative 0.8317333 equals negative K times 7.3 days divide from both sides negative 7.3 days. And we'll get our K. I'm gonna take that K and plug it into my half life equation. The K that I isolate is 0.113936 days in verse. So when I plug that in, I'm gonna get as my number of my half life here as 6.1 days. So 6.1 will be our final answer.

Radioactive Half-Life Example

Now method three is the easiest one to spot because in this method, they're talking about fractions or percentages. So in method three, we also utilize both the radioactive half-life equation and the radioactive integrated rate law. In this case, the questions will ask for the fraction or percentage remaining while involving half life. Here, we're going to say that the fraction of radioisotope is equal to your final concentration divided by your initial concentration. And remember if we're multiplying a fraction by 100 that's will give us a percentage. In this case, the percentage remaining of our radio isotope here. If we take a look at our examples or one example, it says here, the half life of iodine 131 in isotope used in thyro thera is 8.021 days. What fraction of iodine 131 remains in a sample that is estimated to be 6.25 months old? All right. So we can use this flow chart to help us organize our thoughts here. We have our starting position. So the first question we ask ourselves is, do you have the decay constant K? Well, if we look here it doesn't tell us what the K constant K is. So we can't move on to. Yes, we have to move on to. No. Well, if it's a no, then use the radioactive half life equation to help us isolate our decay constant K. So here half life equals Ln two over K. We know what the half life is. With that information, I could find my K if I rearrange this equation, K equals I add two over half life. So the Ln 2/8 0.021 days. So here that would give me 0.08642 days in verse for K, right? So now I know, OK, so then I move on, I use the radioactive integrated rate law which is this and I use it to solve for the initial and final concentrations to find our fraction, which again is equal to my final divided by my initial. Now remember we can rearrange this equation for integrated rate law. Since we're looking for a fraction, it can become Ln of your final concentration but divided by your initial concentration and that'll equal negative KT. Now, here we plug in the information that we know R is negative 0.08642 days in verse. But remember K and T have to have the same units here, KS and day in days in verse. So time needs to be in days. So I'll have to change these months into days. Now, this is gonna be an estimate where we're gonna say that 11 month is equal to about 30 days. Of course, some months are 31 days. February is 28 or 29 days. This is just an average. So once you cancel out, I'll get 187.5 days, which I plug in here. So here when we multiply these two together, that's gonna give me negative 16.20375 which is equal to Ln of my final divided by my initial, which is my fraction. I need to isolate just that fraction itself. So here to isolate just that fraction, I'm gonna take the inverse of the natural log on both sides. That's gonna have the Ln drop for my re my left side. And then if I take the inverse of the natural log of on the right side, it becomes E to this number. And if I were to plug that into my calculator, I'll get as my fraction nine point. And here if we want to do in terms of six figs 36 F or 9.18 times 10 to the negative eight. So this would be my fracture remaining for this particular example. Question.

The half-life of arsenic-74 is about 18 days. If a sample initially contains 5.13 x 10⁴ mg arsenic-74, what mass (in mg) would be left after 80 days?

• A. 2.36 x 10³ mg
• B. 7.02 x 10² mg
• C. 1.43 x 10³ mg
• D. 1.14 x 10⁸ mg

What percentage of carbon – 14 ( t_1/2 = 5715 years) remains in a sample estimated to be 18,315 years old? 

• A. 11.09%
• B. 27.96%
• C. 18.72%
• D. 10.85%