Iodine-131 is a convenient radioisotope to monitor thyroid activity in humans. It is a beta emitter with a half-life of 8.02 days. The thyroid is the only gland in the body that uses iodine. A person undergoing a test of thyroid activity drinks a solution of NaI, in which only a small fraction of the iodide is radioactive. (c) A normal thyroid will take up about 12% of the ingested iodide in a few hours. How long will it take for the radioactive iodide taken up and held by the thyroid to decay to 0.01% of the original amount?

Video transcript

Welcome back everyone. We're told that Samari um 153 is used for palliative bone pain treatment. It has a half life of 1.93 days and decays by beta emission. If the bone of a person undergoing treatment absorbs 46% of the administered summary. Um how long will it take for the concentration of the absorbed summary? Um to decay 2.5% of the original amount. We want to recall that radioactive decay A. K. A nuclear decay follows And sorry, this says DK follows first order kinetics kinetics. So first order kinetics would be defined as the natural log of our concentration of our isotope will say end here at a given time divided by the initial concentration of our isotope, which is set equal to our negative rate constant, multiplied by the amount of time that passes. So in the case of our isotope summary, um 1 53. What we will have is that our initial concentration of sumerian 53 is equal to 530.46. Where we're told from the prompt that 46% is absorbed of treatment. And so for our concentration of our Samari um Once at the case to 5% of the original amount is going to equal .46 times .5%. Which is going to give us a result of 2.3 times 10 to the negative second power. Now we need to figure out what our Rate constant K is going to be. And so we're going to have to recall that since this is following. First order kinetics for our half life. We're going to recall that we have our half life equal to .693 Divided by our rate constant K. and so solving for our rate constant K. We would have .693 divided by our half life. So this first part is our equation that we should recall for first order kinetics half life. We're isolating for K. And sorry our denominator here should be our half life T. One half where we would plug in the half life given from the prompt. So we would have K. Equal to 10.6 93 in the numerator divided by the half life given in the prompt as 1.93 days. Where we will have K. equal to the value of .335-91 inverse days. So now going back to our first order kinetics rate law, recall that we have the natural log of our concentration of so marium at a given time which is then divided by the concentration of. So marium initially equal to negative one times our rate constant K. Times our time that passes. So T. And so plugging in what we know we have the natural log of the value which we determined above as our concentration of Samari. Um After the 5%. That remains from what is originally absorbed. Which is this value here. So we're plugging that below 2.3 times 10 to the negative second power. This is then divided by our concentration of summary um initially which we determined to be 0.46 according to the prompt set equal to our negative rate constant which we determined above as 0.35 inverse days Multiplied by our time which we want to solve for. And so were we would simplify this to the natural log of 0.05 which is set equal to negative one times our half life of 0.3591 inverse days multiplied by T simplifying further. We have negative 2.99573 equal to our right hand side. Negative 0.3591 inverse days times time Where we would keep simplifying to divide both sides by negative 0.35 91 inverse days. So it cancels out on the right hand side. And this would give us the amount of time left equal to a value of 8.34, which we would say is about eight days. And this would be our final answer to complete this example as the amount of time it takes for our solarium to absorb to 5% of its initial amount. So eight days is our final answer. I hope everything I reviewed was clear. If you have any questions, leave them down below and I will see everyone in the next practice video