A sample of F-18 has an initial decay rate of 1.5⨉10⁵/s. How long will it take for the decay rate to fall to 2.5⨉10³/s? (F-18 has a half-life of 1.83 hours.)

Video transcript

Hello everyone in this video which I calculate for the time that it will take for the decay rates to fall to this raid right here. So let's get started. First of all, we're dealing with a radioactive slash nuclear decay of isotopes that follow the first or connects. So the integrated rate law for the first order connections or reactions is going to be as follows. So we have the natural log of the concentration at the time Equalling two negative decay constant multiplied by time and then we're adding our initial concentration. All right. So we can also recall that the half life is a time that's needed to have the amount of reacted to decrease by 30% or one half. So the half life of a first or directions given to us by this equation here then T half equaling two Ln of two overcame. So when you first calculate the decay constant using the given half life of 2.78 hours. So we're basically using this equation and icing for our key. And what we can get out of this is going to be that the Cape is equal to the natural log of two divided by T to the half. So putting in some numerical values then we have Ln of two divided by 2.78 hours. And my final numerical value is going to be 0.2493 hours to negative one. So we don't have a concentration that is given to us. So we can go ahead and use the given decay rate instead of our concentration let's go ahead and sort out what we're given. So what we're given is going to be the concentration at the time being 1.5 times 10 to the third. And then we also have the concentration Or the initial concentration. So that's going to be 2.5 times 10 to the five. Then for a constant will produce software is 0.2493 hours to negative one. And then T. Is what we're solving for. So we do not know this information. Just go ahead and solve for time. So solving for time we have the L. N. Of 1.5 times 10 to the third per second equaling two R negative 0.2493 hours to negative one times T. Plus L. N. Of 2.5 times 10 to the five per second. So simplifying this down more. We have 7.31-2 equaling 2 -0.2493 hours of course is multiplied by T. And then adding our Ln of 2.5 times 10 to the fifth. Alright again simplify this further. We have 0.2493 Hours, multiplied by time. Be going to 12.4292 -7.3132. We're basically just going ahead to isolate our t. And if we do so which is by divide each side by this value here, you will get the numerical value of 20.5 and units being hours. And that's going to be my final answer for this video. Thank you all so much for watching.