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After 135 minutes, her grams = radioactive goo experiment, scientist has 216 At the beginning of a sample haz decayed to 13.5 grams. What is the half-life of the g...

Question

After 135 minutes, her grams = radioactive goo experiment, scientist has 216 At the beginning of a sample haz decayed to 13.5 grams. What is the half-life of the goo in minutes? the amount of go0 remaining at time G() Find formula for G()How many grams of goo will remain after minutes?

After 135 minutes, her grams = radioactive goo experiment, scientist has 216 At the beginning of a sample haz decayed to 13.5 grams. What is the half-life of the goo in minutes? the amount of go0 remaining at time G() Find formula for G() How many grams of goo will remain after minutes?



Answers

Radioactive Decay The half-life of a certain radioactive substance is 65 days. There are 3.5 g present initially.
(a) Express the amount of substance remaining as a function of time $t .$
(b) When will there be less than 1 g remaining?

Okay, in this problem, we're told the half life of a substance is 2.45 minutes, and we're told that the substance decreases from 132.8 grams to 8.3 grams and were asked to figure out how many half lives that is and how long that takes. And I'm going to show you two different ways to do this problem. The first way works because this problem was conveniently chosen with numbers that work out nicely. But the second way works for any problem, even if the numbers don't work out quite as smoothly. Okay, so the easiest way would be to figure out, ah that it takes 2.45 minutes for the substance to go from 132.8 grams down to half of that, which is 66.4. So that's been two minutes at 2.45 minutes. Now it's going to go from 66.4 grams down to half of that, which is 33.2. So that took another 2.45 minutes. And now it's going to go from 33.2 grams to half of that, which is 16.6, and that takes another 2.45 minutes. And then it also finally goes from 16.6 grams down to 8.3 grams, which took another 2.45 minutes. So that worked out to be 4/2 lives for half lives, we could say, And the total amount of time if you add all those together is 8.18 minutes. Now that worked out great, because it was exactly 4/2 lives. But if it wasn't a number that worked out so perfectly, we might want to do it a different way. So here's what I would suggest. Okay? We're going to use our formula for exponential decay, which is why equals a knot. Okay, times e to the Katie. And if we're talking about half the original amount of substance, then for why we're going to use half of a knot a knot can stay a knot and were given the time the half life as 2.45 so we can substitute that in for tea and saw for Kay. All right, so we're going to divide both sides of the equation by a knot. And that leaves us with 1/2 and I'm just gonna write. That is 0.5 equals E to the 2.45 K. And then we're going to take the natural log of both sides. So now we have natural log of 0.5 equals 2.45 k and divide both sides by 2.45 So now we have a value for K. K equals the natural log of 0.5, divided by 2.45 And that is our constant for this particular substance. Now that we have that we go back again to the equation, why equals? Why not or y equals a not each of the Katie. We substituted our final amount for why, which is 8.3 grams and our initial amount for a knot which was 132.8 grams. And we substitute in our value of K, which is natural log of 0.5 over 2.45 and we saw 14. So we divide both sides by 1 32.8 and that gives us 0.6 to 5 equals E to the natural log of 0.5 over 2.45 T. Now, at this point, you might recognize that 0.0.625 is equivalent to 1/8 or one. Let's see, What is it? 1/16 I think it is. And so that might be a clue that you are going to have an even number of half lives. Okay. From here, we're still trying to sell for tea. So we're going to take the natural love of both sides. And we have natural log of 0.6 to 5 equals the natural log of 0.5 over 2.45 times t. And finally, to get tea by itself, we want to multiply by the reciprocal of natural light 0.5 over 2.4 pipe. So, in essence, that means we're going to multiply by 2.45 and divide by natural audible 0.5. And when you plug that hole thing in the calculator, you get 8.18 minutes, and that tells you the total time it takes. And then if you take that and you divide it by 2.45 The amount of time it takes for each half life you'll get, you will get four. So that tells you you have 4/2 lives and it takes 8.18 minutes total. So same answer two different ways to get it.

All right, so in this problem, we are going to be using the half light formula, which is the followings. We've got a equals A not times 1/2 to the tea over H. And so what Those letters all mean is a equals. The final amount a no equals the initial amount t equals the time, and h equals are half life. So we're gonna be taking advantage of this formula throughout all of the problems that we're gonna do here today. So in our first problem, we are looking to figure out an expression for the mass after a certain number of hours. Now we're given an initial amount, but we're not really given much else. So we are going Teoh put in the information that we were given in order to come up with our expression so we don't know our A but we do know are a not so we're gonna put that right here. And remember, toe always put your units. Times are 1/2. We do not know r t, but we were given or half life value. So we're gonna put that right here. So we're just plugging in the values that the problem actually gave us and that sums up all we have to dio for part a. Now we're going to go on to part B, so Part B were actually given a amount of time that they want us to put in. And then we want to find the final value so we actually have an expression all lined up. So we have to do is put in what they gave us. So we have a equals. Our initial mass times are 1/2. Now we have our tea, which in this case is going to be 75 hours over 578. So what we're gonna do here is we're actually gonna put this into the calculator, and we should be good to go. And so once we put it into the calculator, we get a certain value, which is approximately 58.49 And again, we want to put our units here, which in this case, is milligrams. Also, one thing that I would suggest you do is figure out how many Sig figs you want to include in this problem. So I used to hear if you wanted to use one, it could be 10.5. If you wanted to use more, you would extend that value and have more. So just make sure that you know what kind of answer you're expected to give. So for the last part, we are actually given an a value. So we're gonna plug that in here. We're gonna also put in all the other information that we have been given. So the initial value remains the same with 1/2 remains the same. But this time we're not actually given amount of time. We're trying to figure that out. So we're going to leave that as a variable of tea, and we're gonna put in our same half life that we've been using the entire time. So our goal of any algebraic expression is you want to make sure that this t the variable ends up isolated. So our first appears we're gonna divide by 64. There we go. And I'm gonna just scroll ups that we have a little bit more room to work. And so when we have 12/64 we're getting approximately 0.1875 and then we have this expression on the other side but are variable is still an exponents. And so we need to get that exponents by itself. So to do that, we're actually gonna be taking the natural log. So we're gonna take the natural log of both sides. Now, when we have an expression that looks like this whenever we take the natural log or any log we can actually take the exponents and put that in front of the expression. So then it's this times the natural log of 1/2. Now, we still have to do a few more steps in order to get the t by itself. So we're gonna divide by the natural log of 1/2 on both sides. There we go. And that ends up getting us our t almost by itself. So if we actually calculate this value, we would get approximately negative 1.674 over negative 0.693 equals this. So now it looks a lot more like a typical algebraic expression. So we're gonna multiply both sides by the denominators that you can cross this out and actually get t by itself. So then what I got as my official value is approximately 1395.89 hours, and that's how you would solve that

This question does us that the half life off tell him to 08 is 3.1 minutes. Behalf or the half life is equal to 3.1. Mhm. Yeah, and then they're asked, um to calculate the grams off remaining. Tell him to go eight after 12 minutes if there was, ah, mass off 84.6 g being studied in the laboratory. So the initial mass here is, as we just heard is 84.6 grams and were asked to calculate how much of this is going to be remaining after t equals 12 minutes in the first part of the question. This is the data that's available so we can use this'd equation There. We can calculate the mass off Valium remaining at Time T, which is after 12 minutes by taking the mass off pallium that was present at Time zero, which is 84.6 g and multiplying that by 0.5. Just raised to the poverty already have. Can you fee at the values? Uh huh. We did 84.6 g. Multiply that by 0.5. The year is 12 minutes, and the half life is 3.1 minutes. So these are both given in a minute units and they'll cancel out each other. So the answer to this part of the question is going to be equal toe five fine. 78 Gramps, this is the mass off Valium to, oh, age remaining after 12 minutes. In the second part of the question, they asked to calculate something else and that iss how many minutes, uh, will it take for the mask off Italian? Tow it to reduce to stay point for a grams. So let's take a second scenario. Now we have different data in this case, G is equal toe, but we have to calculate. So it's unknown, and the M not is the same as, um for the previous question, because they both began at the same time and began with the same initial mass. So it is still 84 84.6 grams and new mass at Time T is equal to 3.48 g and there are two calculating. So we have this equation we can use to do this calculation. Mm hmm. But we used the version that ISS solved 40. It will look like this D or the time it has bean is going to be equal to the negative half life divided by 0.693 How did you blow this by the natural log off, Pinky over And hope so instead of empty over. And now we can also use the masses empty over and not m d over m or no. If we substitute for these, we get minus three point one minute, which is the half life. Divide this by 0.693 kind of blood to spy the natural log off the mass at 20 which we're told is 3.48 grams. The mass at time zero which at all is 84 point six and then cams Graham units will cancel out each other. And when we solve for this, we get about 14.3 minutes. So after about 14 0.3 minutes, the mass off this radioactive isotope empty is going to reduce 23.48 grams

In this question. We have a radioactive element that has 1/2 life of three hours. So if we start with 400 rounds, so that is going to be are a one were asked, What is the amount after 12 hours? Okay, so because we're talking about half life, that means our common racial are is going to be when his Hey, So we can use this now to find the general terms so they on is going to be a one times larger. The power on minus one, where a one is 400 and are is 1/2. Okay, So now the question is, um what is n if we're talking about 12 hours, some Since the half, life is three hours. We need to calculate how many half life half lives will we go through and 12 outs. So if the half life is three hours than in 12 hours, we would have gone through it four times. Right, Cause every three hours it would, um, you would occur once. So in this case, uh, for 12 hours, we would go through four of them. Okay, So now anyone, That is what happens at, um, time zero. So which means so let's draw a table on the side. So we have our time and now we have our amount. So at times zero it is 400 and then after 1/2 life, so that's going to be three hours. But this is time and ours the what? Diminished the 200 and then after six hours, that's another half life. So that's 100. Okay, so we're looking for 12 hours. So really, this is are a one B two a three a four, a five. So we're actually looking for a five and not a four, because a one is what happens. A one is before any half life has occurred. So we are looking for hey, which is this number right here. And that is going to be 400 times 1/2 the power of I think so. That ends up being 25 grand


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