Radioactive Half-Life - Video Tutorials & Practice Problems
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1
concept
Intro to Radioactive Half-Life
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Here we can say that radioactive half lives, which is T sub half, is just the amount of time required for half of a radioisotope to decay. Now, a radioisotope or nuclei is just an isotope that has an unstable nucleus, and emits radiation as it decays. Alpha decay, beta decay, electron capture, positron emission, etcetera. Based on this, if we take a look here, the example says what is the half life of the radioisotope that shows the following data of remaining percentages versus time. So here we have our percentage remaining, and here we have number of days. We start out with a 100% of our starting material on day 0. Remember, half life looks at how much time it takes us to lose half. Alright 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 0, it takes 3 days for us to lose half, Looks like it takes another 3 days to lose half, and it looks like it look took another 3 days to lose half. So based on this chart here, it looks like the half life is equal to 3 days for this particular radioisotope.
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concept
Method 1 of Radioactive Half-Life
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Now we're going to look at different methods that involve half life in some way. In method 1, we're dealing with the direct calculation of half life or rate constant. Now in method 1, 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 ln2 over k. Ln2 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 through the whole reaction. Remember the example that we saw earlier, every 3 days we lose half of our starting material. It was every 3 days, it didn't fluctuate where it's 2.5 here, and 1.8 here, or 7.2 here. It was always 3 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'd 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.
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example
Radioactive Half-Life Example
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Here we're told if the decay constant of Plutonium 244 is 8.66 times 10 to the negative 9 years at 25 degrees Celsius, what is its half life? Alright. 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 ln2 over k. Plug in the value that we just got, which was 8.66 times 10 to the negative nine years, inverse, should be inverse, that will give me 8.00 times 10 to the 7 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.
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example
Radioactive Half-Life Example
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Now, method 2 deals with radioactive nuclei concentrations. In method 2, 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 naught. 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 to the 4 disintegrations per second. After 7.3 days, its activity now becomes 3.7 times 10 to the 4 disintegrations per second. What is the half life of radon 2/22? Alright. So half life equals ln2 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 3 variables we can solve for our 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 going to say ln of final amount, final concentration equals negative kt plus ln of initial concentration. So here we'd say ln of 3.7 times 10 to the 4 equals negative k which we don't know, we have 7.3 days, plus ln of initial. We're going to subtract ln of r initial from both sides. When we do that here we are going to get initially -0.8317333 equals -kx7.3 days. Divide from both sides negative 7.3 days, and we'll get our k. I'm going to take that k and plug it into my half life equation. The k that I isolate is 0.11 3936 days inverse. So when I plug that in, I'm going to get as my number of my half life here as 6.1 days. So 6.1 will be our final answer.
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example
Radioactive Half-Life Example
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Now, method 3 is the easiest one to spot because in this method they're talking about fractions or percentages. So in method 3, we also utilize both the radioactive half life equation, and the radioactive integrated rate law. In this case the questions will ask for the fractured 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 a 100, that's will give us a percentage, in this case the percentage remaining of our radioisotope. Here if we take a look at our examples, or one example, it says here the half life of iodine 131 and isotope used in thyrotherapy is 8.021 days. What fraction of Iodine 131 remains in a sample that is estimated to be 6.25 months old? Alright. So, we can use this flowchart 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 decay 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 ln2 over k. We know what the half life is, with that information I could find my k. If I rearrange this equation, k equals ln2 over half life. So it would be ln2 over 8.021 days. So here that would give me 0.08642 days inverse for k. Alright. So now I know k. 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 our 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 will equal negative kt. Now here we plug in the information that we know. Our k is negative 0.08642 days inverse. But remember, k and t have to have the same units. Here k is in days inverse, so time needs to be in days. So I'll have to change these months into days. Now this is going to be an estimate where we're going to say that one 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 months here cancel out and I'll get 187.5 days, which I plug in here. So here when we multiply these 2 together that's going to 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 going to take the inverse of the natural log on both sides, that's going to have the l n drop from my left side, and then if I take the inverse of the natural log 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 9 point, and here if we want to do in terms of sig figs, 3 sig figs, so 9.18 times 10 to the negative 8. So this would be my fraction remaining for this particular example question.
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Problem
Problem
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
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Problem
Problem
What percentage of carbon – 14 ( t1/2 = 5715 years) remains in a sample estimated to be 18,315 years old?