The half-life for the radioactive decay of C-14 is 5730 years and is independent of the initial concentration. If a sample of C-14 initially contains 1.5 mmol of C-14, how many millimoles are left after 2255 years?

Video transcript

Welcome back everyone. Tritium has a half life of 12.32 years, which is independent of the initial concentration. Calculate the number of minimal present in a sample after 26.45 years. If the sample initially contained 6.4 millimoles of tritium, let's recall the half life formula for a radio isotope where the half life T one half is equal to the Ln of two over the decay constant K. In order to find the minimal present after 26.45 years of tritium, we're going to need to solve for the decay constant K and plug that into our integrated rate law formula. So isolating both sides for K, we would multiply both sides by the decay constant K. So that, that term cancels out on the right in the denominator. And that will simplify so that we have the half life multiplied by the decay constant equal to the Ln of two. And so to simplify further, we would divide both sides by T one half where we would find that the decay constant K is equal to the Ln of two over the half life. And solving for the decay constant, we would plug in the results of our numerator, which from our calculators would be a value of 0. which is divided by our half life given in the prompt of 12.32 years for tritium. And so simplifying this quotient in our calculators, we would find the decay constant equal to a value of 5.62619 times 10 to the negative second power inverse years. Take note that according to the prompt, we are given that the half life of tritium is not dependent or does not depend on the initial concentration of tritium. And so therefore, this tells us that the decay occurring is a first order decay. So we're going to recall the integrated rate law for a first order decay. And so that would be set up to where we take the Ln of the concentration of tritium at a given time T which is set equal to negative one multiplied by the decay constant K, which is multiplied by the amount of time that passes plus the Ln of the concentration of our radio isotope tritium initially. So plugging in our notes, we have the Ln of tritium at a given time, which is what we are solving for. This is set equal to negative one multiplied by our decay constant, which we determined above to be 5.62619 times 10 to the negative second power inverse years, that is then multiplied by the amount of time that passes which according to the prompt is 26.45 years, making more room in the line that is then added to the end of the concentration of tritium initially, which we are given from the prompt as 6.4 millimoles. So we have the of 6.4 millimoles in our next line, we would simplify so that we combine the product of the year terms. So that in our next line, we have the N of tritium at a given time is equal to the result of that product which is negative 1.4813. And that is added to the result of the Ln of 6.4 which is a value of 1.8563 millimoles. And then in our calculators, taking the sum between these two terms, we have the Ln of tritium at a given time equal to the result of the right hand side which is 0.36817. Getting rid of the Ln term, we're going to make both sides exponents to euler's number. And in our final line that will simplify so that the concentration of tritium after the amount of time of 26.45 years is equal to 1.44509 which we will round to about 36 figs as 1.45 millimoles and sorry. So we should have minimal carried on in our work. And so our final answer is 1.45 millimoles. I hope that this made sense and let us know if you have any questions.