Algebra (all content)
Sal constructs a function to model the decay of a radioactive element. Created by Sal Khan.
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- Isn't the radiological half-life of Cesium-137 approximately 30 years, not 30 days?(75 votes)
- I believe you found a mistake. . . it's 30.17 years according to wikipedia http://en.wikipedia.org/wiki/Caesium-137(41 votes)
- if after every 30 days the cesium-137 has half of what it had before, then it would go on infinitely because it just keeps on halving itself? I think it keeps getting closer and closer to 0 like an asymptote right?(12 votes)
- Theoretically yes, but cesium has a finite subdivision, meaning unlike pure numbers which are split into infinitesimals, cesium is split into a finite integer number of particles. And either way, once a radioactive element reaches a stable element isotope, it will stop decaying.
Nice question though.(13 votes)
- Is there something named after Marie Curie except for the element? Like is there a unit called the Curium that measures something like the Becquerel?(5 votes)
- A Becquerel is only 1 nucleus (decayed) per second. A Curie (originally the radiation from a gram of radium) is now 3.7 *10^10 nuclei per second.(1 vote)
- how much is a becquerel?(4 votes)
- A becquerel (Bq) is a measure of how radioactive something is. It tells you the number of disintegrations per second.(7 votes)
- at0:20sal says that the half life of cesium 137 is 30 years and the weight changes from 2kg to 1kg. so at the end of the half life period there will be only half of the elements mass left, right?!
- Yes, but please understand that the mass just doesn't disappear. The cesium-137 turns into barium-137. So the half of the mass of cesium that decayed didn't just disappear, it merely changed into a different element. However, there is a small amount of matter-energy conversion involved.(3 votes)
- It can be annoyingly difficult reconciling exponential growth / decay problems done with Algebra 2 methodology with those done with differential equations. Are there videos comparing the two methods?(3 votes)
- how am i supposed to write 5^sqrt(0.5)? on the calculator?(2 votes)
- Instead of taking the x-th root from something, you can raise it to the power 1/x.
Square root would be ^(1/2) or ^0.5.
30th root would be ^(1/30).
- In the radioactive decay series, the equation was A(t) = A(0) * e^-lambda*t. This example confuses me.(2 votes)
- I'm right with you on this one. In class we are looking at exponential growth and decay using exponential function N(t)=N(0)e^(kt), where N(0) is the initial amount, k is the growth rate, and t is time. I'm having a difficult time relating this to A=Cr^t.(2 votes)
- would it be correct if we said A(t) = c * (1/2)^(t/30) instead, where t is in days?(2 votes)
- Yes this is equivalent.
And this way you can clearly see that one half decays every 30 days. :)(2 votes)
Caesium-137 is a radioactive tracer element used to study upslope soil erosion and downstream sedimentation. It has a half-life of approximately 30 days. So this half-life of 30 days, this means that, if you were to start with 2 kilograms of caesium-137, that 30 days later you're going to have 1 kilogram of caesium-137. The other kilogram has decayed into other things. And if you waited another 30 days, you would have half a kilogram of caesium-137. Assume that the amount A in becquerels of caesium-137 in a soil sample is given by the exponential function A is equal to c times r to the t-- where t is the number of days since the release of caesium-137 into soil and c and r are unknown constants. So this bears some explaining. What is this becquerel business? So normally, if I were to talk about the amount of some element, I'd probably be thinking in terms of mass and I might talk in terms of kilograms. But some people might also be referring to the amount of this radioactive substance in terms of the amount of radioactivity it produces. And becquerels is the international unit of radioactivity-- named after Henri Becquerel, who codiscovered radioactivity with Marie Curie. So you could consider this the amount of caesium-137 that causes A becquerels of radioactivity. But either way, we can just think of it as a quantity. But it's really the quantity that causes A becquerels of activity. So just to be clear, the amount is given by the exponential function A is equal to-- let me just rewrite it-- A is equal to c times r to the t power-- where t is the number of days since the release of the caesium-137 in the soil and c and r are unknown constants. Fair enough? So let's just be clear, this is days since release. In addition, assume that we know that the initial amount of caesium-137 released in the soil is 8 becquerels. Solve for the unknown constants c and r. So the initial in the soil. That's when t is equal to 0. When no days have passed. So we could say that the amount at times 0-- well, that's going to be equal to c times r to the 0 power, which is just going to be equal to c times 1, which is equal to c. And they tell us what A of 0 is. They say A of 0 is 8 becquerels. So this is going to be equal to 8. So our constant here, the c is just going to be equal to 8. What is the value of the constant? We could just write 8 right over there. So the value of the constant c is 8. What is the value of the constant r? Round to the nearest thousandth. So we're starting with 8. So A of 0 is 8. How much are we going to have after 30 days? And the reason why I'm picking 30 days is-- that is the half-life of caesium-137. So A of 30, remember our t is in-- let me just switch colors just for fun. Remember, t is in days. So A of 30. So after 30 days, I am going to-- if I want to use this formula right over here, if I wanted to use the description of this exponential function, we already know that c is 8. It's going to be 8 times r to the 30th power, which is going to be equal to what? Well, if we started with 8, 30 days later, we're going to have half as much. We're going to have 4 becquerels. And now, we can use this to solve for r. So you have 8 times r to the 30th power is equal to 4. Divide both sides by 8. You get r to the 30th power is equal to 4 over 8-- which is the same thing as 1/2. And then, we can raise both sides to the 1/30th power. r to the 30th-- but then, you could think of the 30th root of that or raising that to the 1/30th power-- that's just going to give us r is equal to one half to the 1/30th power. And that is something that's very hard to compute in your head. So I suggest you use a calculator for that. And they hint because we're going to round to the nearest thousandth. So let's get a calculator right out. And so we're talking about one half to the 1/30 power. So we get 0.9771599-- it keeps going. But they tell us to round to the nearest thousandth, so 0.977 rounded to the nearest thousandth. And then, they finally say-- how many becquerels of caesium-137 remain in our sample 150 days after its release in the soil? Use the rounded value of r, and round this number to the nearest hundredth. So just to be clear, we already know c and r. We know that the amount of caesium-137 in becquerels-- as a function of time in days-- is going to be equal to 8 times 0.977 to the t power-- where t is the number of days that have passed. And they're essentially saying, well, how much do we have left after 150 days? So they want us to calculate what is A of 150? Well, that's going to be 8 times 0.977 to the 150th power. And so clearly, we need a calculator for this. So let's calculate that. So it's going to be 8 times-- and they tell us to use our rounded value of r, not the exact value of r. So it's going to be 8 times 0.977 to the 150th power. And they want us to round to the nearest hundredth. 0.24. 0.24 becquerels is kind of the radioactivity level of the caesium-137 that we have left over. Now, one interesting thing is they asked us to use the rounded value of r. So we used the rounded value of r. Because this right over here is a multiple of 30, you could actually-- in not too difficult of a way-- find out the exact value that's left over. And actually, you don't even need a calculator for it. I encourage you to pause your video and try to think about that. Find the exact value. Well, instead of writing 0.977, let's write A of t as being equal to 8 times our r. This is an approximate value for r. If we wanted to be a little more exact, we can say that our r is one half to the 1/30th power. And we're going to raise that to the t power. Or we could say A of t is equal to 8 times one half to the t/30 power. If we raise something to an exponent and then raise that to an exponent, we can take the product of those exponents. So that's one half to the t/30 power. Let me actually do that in another color. Let me do that in yellow. So that's 8 times one half to the t/30 power. And actually, I don't need this parentheses right over here. This is another way to describe A of t. So what is A of 150? So A of 150 is going to be equal to 8 times one half to the 150/30. Well, that's just 5. One half to the fifth power. Well, what's one half to the fifth power? That is 1 to the fifth over 2 to the fifth or 1/32. So this right over here is 1/32, which is equal to 8 over 32, which is equal to 1 over 4, which is equal to 1 over 4 or 0.25. So using our approximation for r, we got 0.24 when we rounded to the 150th power. So that's using our approximation a lot. We're taking 150 of these and multiplying them together, but it's not too far off of what the real value is. And they asked us to use the rounded value. But if we'd used the precise value, the actual value, 0.25 becquerels would be left over.