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~/2 POINTSLARAPCALC9 4.6.012.Complete the table for the radioactive isotope: (Round your answers to two decimal places. Half-life Initial Amount after Amount after ...

Question

~/2 POINTSLARAPCALC9 4.6.012.Complete the table for the radioactive isotope: (Round your answers to two decimal places. Half-life Initial Amount after Amount after Isotope (in years) quantity 1000 years 10,000 years226RA1599grams1.3 gramsgrams-/1 POINTSLARAPCALC9 5.1.048Find the particular solution that satisfies the differential equation and initial conditionf"'(x)X > 0; f(1) = 1f(x)POINTSLARAPCALC9 5.3.020.o Kai conctant of Intearation. Remember to use absolute values

~/2 POINTS LARAPCALC9 4.6.012. Complete the table for the radioactive isotope: (Round your answers to two decimal places. Half-life Initial Amount after Amount after Isotope (in years) quantity 1000 years 10,000 years 226RA 1599 grams 1.3 grams grams -/1 POINTS LARAPCALC9 5.1.048 Find the particular solution that satisfies the differential equation and initial condition f"'(x) X > 0; f(1) = 1 f(x) POINTS LARAPCALC9 5.3.020. o Kai conctant of Intearation. Remember to use absolute values



Answers

Complete the table for the radioactive isotope. Amount After I000 Years $1.5 \mathrm{g}$ $0.4 \mathrm{g}$ Initial Quantity $10 \mathrm{g}$ $3 \mathrm{g}$ Half-Life (years) 1599 1599 5700 24,100 Isotope Complete the table for the radioactive isotope. Amount After I000 Years $1.5 \mathrm{g}$ $0.4 \mathrm{g}$ Initial Quantity $10 \mathrm{g}$ $3 \mathrm{g}$ Half-Life (years) 1599 1599 5700 24,100 Isotope $$^{226} \mathrm{Ra}$$

Okay, so let's solve this problem where Q represents a mass, the mass and grams of carbon 14 after so many years, t years. So our first job is to determine the initial quantity. So once he is equal to zero, we can do that in our heads honestly to find Q when T. Is equal to zero. Yes. And I say that we can do it in our heads. Because if you just take a look at this function when T. Is zero, this is going to be 10 times one half to 20 1/2 to the zero is just one. So that leaves us with Q Equals 10 g. This is going to be our starting amount. Okay, part B heart be We're going to determine the quantity present after 2000 years. So that means cute when T equals two 1000. So let's put that in Q is equal, going to equal 10 times one half to the Sorry I should say 2000 Divided by 50 700. So all we have to do here is plug this into our calculators. And we will get when we evaluate this 7.84 g. 7.84 grams. Okay then part C. A sketch a graph of the function Over the time period zero to 10,000. Okay. So if we have our tea access here, which is years. Okay. And our cue access here, which is mass of carbon 14, then our zero point here on the keep you access Is going to be our starting point when two U 0. So 10. And then if we have 10,000 all the way down here In 2000 will fit somewhere here. 2000. We know that for 2000 we have seven points 8 4. So it's going to be I'm just guesstimating a little bit. Yeah, something like that. So then we draw this graph. it's going to look like fact. Okay, that looks a little straighter than I'd like it to look because remember this is an exponential decay function. So I'm just gonna keep driving this until it turns out a little bit better. Okay, so there's our function, it's not a straight line. If I zoomed out a little bit more and it was a better artist, it looked more like that. So here let me just rework this seven seven point, it's going to be a little lower. So we'll make the drawing look a little better. So then there we go, that's better. Okay, So our first step is to find the initial by plugging in T equals zero. And then after 2000 years plug in 2030. And then we sketched this graph and we know it's exponential decay because that's what half life problems are. And because we have these 2.010 and 2007.84, so there we go

All right. So, we're finding the exponential decay. We have the half life the initial amount and we aren't sure what the amount will be after 1000 years. So we can use this half life and set it up. All right. So one half when we go to e. Next race the negative K. Trying to find. Kay times our time which we've gotten to be 1 1599. So, we can take the natural log of both sides of this equation. Get the natural log of 1/2 Is equal to negative 1599 K. We'll get that K. Will be equal to Natural log of 1/2. Bye Bye. Negative 1599. So then we can find K. two b. Very small number. Okay, so our equation now will be a equals 10 g. Let's say. Alright. Alright. R. K. value here. It's going to 0.000 for Times are time. And we're finding it for 1000. Like this year, we're going to put in 1000. So then we can solve for a of 1000. She lives with 6.5 g remaining After 1000 years.

Question 21 wants you to complete the table for the radioactive isotope. 2 26 are a to do. So you want to the formula A equals P E to the R T where a is your amount of that isotope at time? T he is your initial amount. Initial quantity T is time in years and our is your rate of decay. And you wanna make sure that's in decimal form when you plug it in. Given in question 21 is your half life in years is 500 or 1599 years. Your initial quantity P is equal to 10 g, and we want to find how much at 1000 years. So starting off we know that are half life is 1599 years, which means that the ratio from the amount that we have left due to the initial quality or over the initial quantity, is equal to one half. And you can set that equal thio e to we are t um, or tea is gonna be 1599. Then you can take Ln of both sides. Um, and in doing So then you have Ln of one half is equal. 2 15 99 are from there. You know that your rate of decay is gonna be negative. 0.43348 Then you can use R T equals 1000 years, which is what we're trying to find to fill in the table. Yeah, into that formula, where you get a equals 10, which is our initial quantity given e to the negative 0.433 for eight times 1000 and you get A is equal to 6.482 44 g is how much is left after 1000 years, and that's your final answer.

All right. So we have our half life and our amount. After 1000 years. We're trying to find our P. So first things first we gonna find our K. Value and all we need to do that is our half life. We know that one half of the substance will be left the sequel to the race to the negative K. Times our half life period. So to solve for cable and take the natural log of both sides of the equation. Natural 0.5 Is equal to -24 100 K. It's okay. We're equal Natural Law. 2.5 Over negative 24,100 to this will be extremely small number. Okay So we have Okay is equal to 0.000 03 Okay. Mhm. So for our equation we're just trying to find peace so we can set up Are a value here. 0.4 g is equal to P. E. raised to the negative. 0.00003 Time is 1000 years. And we just solve for P. So it will just be he will be equal to her walk. All this off. It would just be that divided by that will be equal to P. So to put that into our calculator. So actually be remarkably close to the same amount. It will be 0.41 g 1000 years. Isn't long for that to decay. 0.41 g will be remaining After 1000 years.


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