A proposed nuclear theory suggests that the relative abun-dances of the uranium isotopes 235U and 238U were approximately equal at the time they were formed. Today, the observed ratio of these isotopes is 7.25 x 10^-3. Give that the half-lives for radioactive decay are 7.04 x 19^8 are y for and 4.47 x 10^9 y for , calculate the age of the elements.

Video transcript

Welcome back, everyone. We're told that Americium is a synthetic radioactive element. And isotopic mixture was made in a laboratory that had equal amounts of 2 41 americium and 2 43 americium in it. After some time, the observed ratio of the two isotopes was 0.1 58. The half lives of 2 41 americium into 43 americium are 32.2 hours and 7370 hours respectively. How old is the mixture after this change? Alright. We're going to begin by recalling that whenever we have radioactive decay, this will correspond to being first order. And so we want to recall the following rate law Where we would take the natural log of N over N0, which is set equal to negative one times the decay constant K multiplied by the time T where we want to note that N represents our Adams of our americium isotope at a certain time being too And, and zero represents our atoms of our americium isotopes when the time is equal to zero. So before any decay occurs, we also want to make note of the fact that we can find K by taking the Ln of two over the given half life in the prompt. And that's going to be our first step. We want to begin by assuming that we have for a merry cm 2 41 a decay constant being K one where we would take the Ln of two being and we'll actually just write the result. It's going to be 0.6 93 in the numerator divided by its half life as mentioned in the prompt being 432.2 hours. And then we would also assume that for a merry cm to 43 we would have a decay constant, a second decay constant rather K two. Where we take again, Ln of two being 0.6 93 divided by the given half life of Americium 2 43 in the prompt being 7370 hours. And so the results of K one R decay constant for Americium 41 we would have a value of 0.160342 inverse hours. Then for Americium 2 43 K two, we have a decay constant, a second decay constant of 9. times 10 to the negative fifth power in verse hours. Now we can go into finding the age of the mixture to solve our prompt. So the mixture between Americium 2 43 and 2 41. And so we're solving for T and so going into our defined rate law, we would have the L N where we take and one for Americium 2 41 over and 01 for Americium 2 added to K one times time. And we're going to set this equal to the same thing for Americium 2 43. So in this case, we would have L N and two over N zero to plus K two times T. And so we set, we reformatted our rate law so that we could set up an equation. So that for both of our isotopes, we have an equal sign between the rate loss. So we're going to begin now by combining our like terms. So in this case, I'll underline them in different colors. We have the L N of N two over N 02, which is like with Ln of N one over N 01. And then we have K one T, which is like with K two T. And so combining these like terms, we're going to subtract the Ln of N two over N 02 so that we can have it on the left hand side. And then we're going to add and actually let me just make sure the color coding is correct. So we used green, so minus Ln and two over end 02 on both sides. And then we would have minus K one T on both sides. So it's going to cancel on the left hand side and Ln of N two over and 02 is going to cancel on the right hand side. And so now we will have our like terms combined so that we have an equation that reads the L N of N one over N 01 subtracted from the Ln of N two over N zero to equal to K two T minus K one T. And so now we can factor out T and L N. And so for the left hand side factoring out Ellen, we would have the L N, then we would have a brace symbol which separates the following quotient where we would have N one over N 01. Now divided by N two over N 02. And this should have parentheses around for the entire quotient which is surrounded by a second base on the other side. Now notice that we divided by and two over N 02 because it was attached by an L in turn being multiplied here. And so this is why we divided, but now we're going to set this equal to on the right hand side where we factor out T, so we would just have K two minus K one in parentheses times T. Now we're going to note from the prompt that in our mixture of Americium 41 And Americium 2 43, we have equal amounts. And so with this in mind, we can say that they're four and 01 is going to equal and 02. And so when the time is zero, before any decay occurs, the initial amounts of our radio isotopes are equal to one another. And so we can simplify even further with this note in mind to say that now we have the Ln of just N one over N two, since we know that N 01 is equal to N 02. And this is equal to our right hand side, which is still K two minus K one times T. And so now we're going to make a second note where from the prompt were given our isotopic ratio after time passes and it's equal to a value of 0.1 58. And so we would plug that in for N one over N to. So instead we would have the L N of 0.1 58. And this is equal to again, K two minus K one times T. And so at this point, because we know our values for K two and K one, we can just plug them in because that was our first step that we did, we solved for K one and K two above. So we're going to plug these values in. And what we'll have is on the left hand side, when we take the LNM 0.1 58 we get a value of negative 1.84516025 set equal to our difference between K two, which we found to be 9.4 to 085 times 10 to the negative fifth power in verse hours. And then our value for K one, which in our first step, we found to be 0. inverse hours. This is all multiplied by time still. So simplifying the right hand side for the difference, we would have the same left hand side being negative 1.84516025. And on the right hand side, the difference simplifies to a value of negative 0. in verse hours multiplied by time T. And so now isolating for T so that we can find the age of our mixture, we're going to divide both sides by negative 0.150939015 in verse hours. It cancels out on the right. And now we would have T isolated for time so that the time or age of the mixture is equal to a value of 1,222. hours. Note that ours is no longer inverse because we are no longer in a fraction or in the denominator rather. And so recall that from our pumped, we have a minimum of three, sorry of three sig figs. And so we would round this to about hours. And so this would represent the age of our mixture of Americium, 2 41 plus Americium 2 43 isotopes. And this will be our final answer which will correspond to choice d in the multiple choice. I hope that everything I went through was clear. If you have any questions, please leave them down below and I'll see everyone in the next practice video.