Rate of Radioactive Decay - Video Tutorials & Practice Problems
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1
concept
Radioactive Integrated Rate Law
Video duration:
3m
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Now recall under chemical kinetics that all radioactive processes or reactions follow a first order rate law. Here we're gonna say for these reactions, we use the following equation. So here we're going to have our radioactive integrated rate law. Here we have our final reactant concentration of a radioactive nuclei. Here instead of using our typical a value we're using N and subzero or sub not equals our initial reactant concentration of a radioactive nuclei. K is no longer the rate constant. It now becomes the decay constant but it is in still, it's still in times and verse time. Here could be in days, years, months, et cetera. But remember the units that time uses is based on the units of K. So if K were in days inverse, that mean that time would have to be in days, always make sure they match together these variables. Give us our radioactive integrated rate law which is Ln, which is our natural log of the final concentration equals negative KT plus allen of our initial concentration. Now here the word concentration doesn't only mean molarity in our earlier sections. Remember when it comes to radioactive processes, we could talk about disintegrations per second. We could talk about a lot of different terminology. It doesn't only mean molarity. Now, this equation is also related to the equation of a straight line. It's Y equals MX plus B where Y is Ln of our final concentration. M here is our slope which is equal to KX is equal to time and B is equal to len of our initial concentration. So remember because it's negative K, that means our slope is decreasing over time. It's negative. With this, we can also think of the plot of Y versus X which again is Ellen of our final concentration versus time. Here. On our Y axis, we have Ellen of our final concentration of our reactant. And then on our X axis, we have time, our initial starts here and the K, our slope is negative. So it's descending over time, slope is equal to negative K. I remember our slope is equal to change and rise over change and run which is delta Y over delta X. This translates to the change in the natural log of the concentration of a reactant divided by change in time. This is just a way of us plotting the information in terms of a graph. But the most important part of this entire section is remembering this radioactive integrated rate law, you'll be employing this equation more often than not compared to everything else we're seeing here. So just remember the variables involved with each one and what they stand for.
2
example
Rate of Radioactive Decay Example
Video duration:
2m
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Here, we're told that the radioactive element of ace 210 has a decay constant of 0.086 hours. Inverse. How many minutes would it take for its concentration to go from 9.3 times 10 to the five disintegrations per second? 22.7 times 10 to the four disintegrations per second. All right. So they want us to find minutes. So they're asking us to find time. Our variable T is the missing variable. All right. So we're gonna have our radioactive integrated rate law which remember would be Ln of our final concentration. Remember concentration here doesn't necessarily mean molarity. In this case, it's in disintegrations per second and that's fine minus KT plus all N of our initial concentration. Remember, your final concentration should be the smaller total because we're decreasing our amount of reactant over time as it's decaying away. So Ln of 2.7 times 10 to the four disintegrations per second RDK constant K we're told is negative 0.086 hours in verse, we don't know what tea is. That's what we need to find plus Ln of our initial, which is the larger amount subtract Ln of 9.3 times 10 to the five from both sides. When we do that, we're going to get initially negative 3.53934772 equals negative 0.086 hours. Inverse times T divide both sides by your K value. And we, it's important to keep the units around. So we can see what our units will be at the end at the end time is in hours. But remember I want the answer in minutes. So we have to do one last conversion. Remember that one hour is equal to 60 minutes. So time here would be equal to 2.5 times 10 to the three minutes. This would be our final answer.
3
Problem
Problem
For the radioactive decay of lead-202 the decay constant is 1.32 x 10-5 yr-1. How long will it take in hours to decrease to 53% of its initial amount?
A
5.72 x 104 hrs
B
5.01 x 109 hrs
C
4.81 x 104 hrs
D
4.21 x 108 hrs
4
Problem
Problem
During World War I radium-226 was used in the manufacturing of luminous paint. If it takes 2.12 x 104 days for its degradation to be 2.49% complete, what is its decay constant?
A
1.74 x 10–4 days–1
B
1.08 x 10–5 days–1
C
1.19 x 10–6 days–1
D
2.14 x 10–5 days–1
5
Problem
Problem
If the decay constant for polonium-209 is 6.80 x 10-3 ys-1, what fraction of it remains after 1.1 x 104 years?
A
9.00 x 10–30
B
3.27 x 10–33
C
8.24 x 10–42
D
9.13 x 10–43
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