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Ch.21 - Nuclear Chemistry

Chapter 21, Problem 35

Some watch dials are coated with a phosphor, like ZnS, and a polymer in which some of the 1H atoms have been replaced by 3H atoms, tritium. The phosphor emits light when struck by the beta particle from the tritium decay, causing the dials to glow in the dark. The half-life of tritium is 12.3 yr. If the light given off is assumed to be directly proportional to the amount of tritium, by how much will a dial be dimmed in a watch that is 50 yr old?

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Welcome back everyone in this example, we have radium in luminescent paints that were used to paint watch faces eventually replaced with Prometheus 1 47 by the second half of the 20th century, we're told that it's a low energy beta emitter. Unlike radium, which is an alpha emitter. Additionally, unlike radium, it doesn't emit penetrating gamma rays. The half life of Prometheus 1 47 is 2.62 years, which makes them easier to dispose of assuming that the light given off is directly proportional to the amount of prometheus determine how much will face of a 15 year old watch be dimmed. So what we're looking for is a percentage of how much of these photons dim to. And what we are going to recall is that because this is an example based on radioactive and nuclear decay, we're going to assume that we follow first order kinetics for the decay of Promethean 1 47. And recall that that law is written out as the natural or the sorry, the natural log of our concentration of our Prometheus um represented by P under case M. At a given time divided by our concentration of Prometheus um initially which is set equal to negative one times our rate constant K. Times the amount of time T that passes where we also want to recall that the half life formula that is used t one half is equal to 10.6 93 divided by our rate constant K. Our first step is to figure out what this rate constant value is based on the given half life in the prompt. So isolating for rate constant, we would have K equal to .693 divided by our half life given as 2.62 years. And this is going to result in a half life or sorry, a rate constant rather equal to the value of .2645 in verse years. So for our final answer, as we stated, we want the percentage of photons that will be dimmed based on our decay of Promethean 147. So we're going to need the value of this quotient here. And so what we're going to do is plug in what we know into this formula. So we would have the natural log of our concentration of Prometheus at a given time divided by the concentration of Prometheus initially set equal to negative one times our rate constant, which we just determined to be 0.2645 inverse years, multiplied by the amount of time that passes according to the prompt being years. So simplifying this, we would have that the natural log of our concentration of Prometheus at a given time divided by the concentration of Prometheus initially is equal to the product here, which will result in negative 3.9676. And to get rid of that Ln term, we're going to make both sides and exponents to Ulis number E which will cancel out Ln here and on the right hand side, we're going to have our concentration of Prometheus at a given time divided by a concentration of Prometheus um initially and sorry that's here equal to Ulis number two, negative 3.9676 equal to the value of 0.2, which will correspond to about 2%. and so we would say that 2% of Promethean which will remain In the 15 year old watch, Meaning that once it was at its brightest capacity being at 100%, it's going to dim By 98%. So it dims by 98%. And so our final answer is going to be the percentage that this watch dims which is by 98%. So this 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.