Fluorine-18 undergoes positron emission with a half-life of 1.10 x 10^2 minutes. If a patient is given a 250 mg dose for a PET scan, how long will it take for the amount of fluorine-18 to drop to 75 mg? (a) 56 minutes (b) 96 minutes (c) 132 minutes (d) 191 minutes

Video transcript

Alright. So in this practice problem we have cobalt 58 which is a synthetic synthetic radio isotope of cobalt with half life of 70.87 days. So we have half life of this isotope. We want to know how many days. So time it will take 49.750 g of this um of this sample of cobalt to finally two decade to final mass of 2.125g. Alright, so we do know that any type of radioactive process is going to happen under first order. Right? So this is the first order. Okay, so we do know which integrated rate law equation we're going to use now we're going to be using two of them actually because if you take a look at the first one, which we actually need. So we have the integrated rate law for first order is going to be a lot of concentration of our compound after a certain time is going to equal to negative negative K, etc. Plus a lot of concentration initially. Okay, so what is given to us well, time here is what we want to find. Right? So this is what we want to find. Initial. We actually have so we have nine points and 50 g initially. So we have this and then we also have the final mass. We have the final concentration. Okay, now we actually don't have K. Which is the decay constant or rate constant. Now, but we do know that it's first order and we know it's half life and the half life equation for first order. It's very simple. It's just half life equals to a line of two divided by K. Which is that the rate constant that we're missing here. Okay. So we're gonna go ahead and solve for K. And then once we have K, we're going to plug it into here and solve for the time. So let's go ahead and do that. So we're going to divide a land of two by the half life that we have. Okay, so, Ellen of two is basically a constant because it's going to give you the same um answer every time you put that into the calculator and then half life is 70. seven days. Okay? So, okay, After you divide that, it's going to be 9.78 mm hmm 9.78 eight times 10. And it's to the negative three. And of course here we have days as the units, but because it is on the denominator, it's going to be days inverse. Okay, so that's okay, we have our case so now we can go ahead and come back here and plug that in and then find time. Alright, so we're gonna go ahead and start working on the situation now on the left and we're going to go ahead and start plugging in the numbers. So we have a land of this is going to be the final concentration, which is in mass, which is fine. So it is 2.125. Okay, that's going to equal to negative case. Okay, here is this number right here, let's go ahead and plug that in 9. To the negative three days universe. And then at times time which we don't know this is our unknown. We're going to solve for that. And then plus the length of the initial concentration which is nine 0.7 50. Alright, so now we're just going to be doing algebra, we're going to be multiplying um adding and all that good stuff. So a lot of 2.125 minus a line of 9. 50 should give you a negative number. 1.52 35. Okay. And then on on the side we're still going to have that negative 9.78 gemstone. Mhm. To the negative three days universe. And that time. So here we still have time we want to go ahead and isolate that. So we're gonna divide both sides by negative K. So we have a negative number here divided by negative. That's going to be positive. Right? So t here obviously time should be positive. Once you divide those numbers, you're going to get one 55.8 where you should get one. Okay, and the time here is going to be in days because here for the rate constant for the decay constant, we had a unit of days universe. But once we divided a number by days in verse that actually comes up onto the top to the numerator, and that's why the time is just in days. Alright, so here's our final answer. This is the time that it would take for that Cobalt 58 to decay from nine g to two g, basically. Alright folks, we're all done. I hope we're able to follow along. If you have any questions, please let us know.