A radioactive sample contains 1.55 g of an isotope with a halflife of 3.8 days. What mass of the isotope remains after 5.5 days?

Video transcript

Hello everyone. So in this video we're trying to calculate for the mass that will remain after 6.4 days in g. So let's go ahead and get started. So we're dealing with radioactive and nuclear decay of isotopes. So that's going to always follow the first order kinetics. So the integrated rate law for the first order reactions is going to be as follows. So we have the natural log of the concentration of time. You're going to the negative decay constant times time plus the natural log of our initial concentration. Alright. So we can go ahead and recall that the half life is the time is needed for the amount of reacting to decrease by 50% or one half. So the half life a first order reaction is given to us by this equation. The T to the half is equal to the Ln of two over K. We need go ahead first calculate for the decay constant. Using the information of the given half life of 4.2 days. So going ahead to manipulate this equation by isolate our K. value. Then we'll get the new value of K. is going to the natural log of two divided by the time. Let's go ahead and plug in some values. That time is going to be 4.2 days. Okay, so putting that into my calculator, I'll get the final value of 0. days to the -1. Since no concentration is given to us we can use the given mass instead of the concentration actually. So let's go ahead and sort out what we are given to the information that we're given. We don't know the concentration at the time. That's something that we need to go ahead and software. But we do know the initial concentration which is or the initial mass in our case Of 2.5 or 2.15 g. As for our concern that we just saw for is equal to 0.165 days. And then our T value is equal to 6.4 days. So now we can go ahead and actually calculate for time. So we have the natural log of the of the concentration is given to us at that time Equal to negative 0.165 days Times 6.4 days. And we go ahead and add the natural log of 2.15. Alright, simplifying this. We have negative 1.056 plus 0.765. Simplifying this even further than Who gets negative 0.291. And now we can go ahead and embrace your power of eve for both sides. And we can get that the concentration at the given time sequel to E. To the negative 0.291. And putting that into my calculator after the final value of 0.75 g. And that is going to be my final answer for this question. Thank you all so much for watching