The half-life for the radioactive decay of U-238 is 4.5 billion years and is independent of initial concentration. If a sample of U-238 initially contained 1.5⨉10¹⁸ atoms when the universe was formed 13.8 billion years ago, how many U-238 atoms does it contain today?

Video transcript

Welcome back everyone. The synthetic isotope of tungsten 1 85 has a half life of 75. days. That does not depend on the initial concentration. A sample of tungsten 1 85 contains 1.38 times 10 to the 19th power atoms calculate the number of tungsten 1, 85 atoms present after three years. Let's begin by recalling the relationship where our half life of tungsten is equal to the natural log of two, divided by our rate constant K. Now we're not given our value for K. So we need to determine what that is. And so we're going to multiply both sides by our denominator K. So that it cancels out on the right hand side. And then we'll have that are half life times K. Is equal to the natural log of two in which to isolate K. We're going to divide both sides by our half life. And so this will cancel on the left hand side and we'll have K isolated so that we find the Ln of two divided by a half life is equal decay. And so plugging in what we know, we'll get four calculators for the results of the Ln of two. A value of 0.693147. And are given half life for tungsten 1 85 from the prompt as 75. days. And so simplifying this quotient in our calculators. We have a result of 9.22966 times 10 to the negative third power inverse days. Next we need to note that are given time that passes in the decay is given as three years. But our decay or rather the rate constant that we just calculated is in units of inverse days. So we want to make sure that this time years is converted two days. So we're gonna take our time equal to three years And multiplied by a conversion factor to go from years in the denominator, two days in the numerator. By recalling that in one year we have an equivalent of 365.25 days. So canceling out our units of years were left with days as our unit of time. And this is going to equal a value of 1,095.75 days. Now note that from the prompt, we're told that our half life of Tungsten 185 does not depend on its initial concentration. And so therefore we would understand that the reaction occurring is going to follow zeroth order kinetics, meaning we want to recognize the following integrated rate law In which we relate the Ln of our concentration of Tungsten 185 at a given time Equal to negative one times our rate constant. K times the amount of time that passes plus the natural log of our concentration of Tungsten 185 initially. And so plugging in what we know, we want to solve for the left hand side. So that would remain the same. And then this is set equal to negative one times our rate constant, which we solved above as 9.22966 times 10 to the negative third power inverse days Multiplied by the amount of time that passes. Given in the prompt as 1,095.75 days. And just to make more room, I'm gonna move things over. So then we can add our natural log of our initial concentration of Tungsten 185 given in the prompt as 1.38 times 10 to the 19th power atoms. And so simplifying so that we can take first our product of the two terms on the right hand side. We would have that our natural log of Tungsten At a given time is equal to the result of that product being a value of negative 10.1134 added to. We have the Ln of our initial concentration of treaty in 1 85 which results in a value of 44.712 atoms of tungsten. Notice that our units of inverse days cancel out with days here. And so we don't have any units after we take that product. Now we're going to continue and take the sum of these two numbers so that we have our Ln of tungsten 85 at a given time equal to the sum on the right hand side, which comes out to be 33. 9578 atoms of tungsten. And so now we just need to get rid of the natural log term by taking both sides as exponents to Mueller's number. That will cancel L. N. On the left hand side. And what we'll find is that our concentration of tungsten 1 85. After the amount of time mentioned in the prompt is equal to 5.59352 times 10 to the 14th power atoms in which we want to round to our lowest number of sig figs being 366 based on the information in the prompt so that we have our final answer of 5.59 times 10 to the 14th power Atoms of Tungsten left after three years. And so what's highlighted in yellow here represents our final answer corresponding to choice c. And the multiple choice. So I hope that this was helpful and let us know if you have any questions