One of the nuclides in spent nuclear fuel is U-235, an alpha emitter with a half-life of 703 million years. How long will it take for the amount of U-235 to reach 10.0% of its initial amount?

Video transcript

Hello. Everyone in this video we try to calculate for the time that it will take for the amount of this isotope to go ahead and reach 25% of its initial amount. So let's first recall that whenever we're dealing with radioactive and nuclear decay of isotopes, it follows the first for directions. So the integrated rate law for first order reactions is going to be as follows. So we have the natural log of the concentration at that time equaling to the negative decay constant plus are multiplied by our time plus the natural log of our initial concentration. All right, we go ahead and also recall that the half life is the time needed for the amount of decreased by 50% or one half the equation. I will always right, will also write down to two half equaling to the natural log of two over K. So we first calculate the decay constant by using the given half life to us which is 34.7 million years. So then we have 34. million years. That's gonna be go ahead and equal to 3.47 times 27 year. Now, kind of messing with this just usa for R. K. Value then. Alright, so we have K equaling to the natural log of 2/2 to the half. Then we have K. equaling to the natural log of two, divided by 3.47 times 10 to the seven year. Alright, now finally put all this into my calculator. I get the constant value. It's okay being 1. times 10 to negative eight years. And because in this problem there's no given concentration. We can go ahead and use the given percentages instead of concentrations. Alright so the information that we're given, we've given the concentration at the time or the percentage at the time we're going to 25%. We go ahead. Actually assume the initial is going to be at 100% R. K. Constantly just solve for is 1.9975 times 10 to the negative eight minutes major over two negative one and for a T. Value we just do not know I was trying to solve for that too. Now go ahead and solve for time. So we have the L end of 25.2% equaling to the negative 1.9975 times 10 to the negative eight Multiply bar time and adding that 100% now. Alright so Again putting in some new record values, we have 3.2189 going to negative 1.9975 times 10 to the -8. Again multiplying this by T. go ahead and add edges with 4.6052. All right now what we're gonna do is again simplify some more. So we have our 1.9975 Times 10 to negative eight years Multiply this by time you're going to 4. -3.2189. So what we're gonna do is just isolate our teeth, which is right over here. So we do that. Look at the value of 6.94 times 27 years, and that's great to meet our final answer first question. Thank you.