The half-life for the radioactive decay of C-14 is 5730 years and is independent of the initial concentration. How long does it take for 25% of the C-14 atoms in a sample of C-14 to decay?

Video transcript

hey everyone in this example we're told that the half life of plutonium 2 39 is independent of the initial concentration. So this tells us that this is a first order reaction. We're also told that it has a value for the half life of 24,110 years, assuming that we start with 100% of plutonium to 39. We need to calculate the time that it will take for 38% of plutonium to 39 sample to decay. So because we recognize that according to that phrase independent of the initial concentration or half life is independent. We would recognize that we have to recall our half life for a first order reaction, which we can recall is calculated by taking the natural log of two, divided by a rate constant K. And so to solve for a rate constant K. We would say that we would take the natural log of two divided by our half life. And so what we would have is that our natural log of two is divided by a half life given in the prompt as 24,110 years. And so this would give us our rate constant K. equal to a value of 2.8749 times 10 to the negative fifth power inverse years. So because we also know that we're following a first order reaction, we're going to recall our first order integrated rate law, which we recall is calculated by taking the natural log of our initial concentration of our plutonium to 39. And sorry, this would be of our concentration of plutonium 2, 39 at a given time. And setting that equal to negative one times our rate constant K. Times time added to the natural log of our initial concentration of plutonium to 39. Now, according to the prompt, we're told that 38% of the plutonium to 39 d. case. And so if we take 100 -38%,, This would give us a difference of 62% of our plutonium to 39. That is left. So plugging this info into our integrated rate law, we would have that our natural log of our 62 of Plutonium 2 39. That is left Is going to be set equal to negative one times our rate constant. Which above we found as 2.8749 times 10 to the negative fifth power inverse years. This is then multiplied by time, which is what we're solving for. And then added to the natural log of 100 Of our plutonium to 39 present initially. So we don't really need to include that initial symbol. So next we would go ahead and solve for T. And what we would get is that the amount of time that it will take for plutonium to 39 to decay will be equal to 1.66 times 10 to the fourth power years. Which we can round to about two sig figs. So we have 1.7 times 10 to the fourth power years. And so this will be our final answer to complete this example. So I hope that everything I explained was clear. But if you have any questions, please leave them down below. Otherwise, I will see everyone in the next practice video.