The age of an igneous rock that has solidified from magma can be found by analyzing the amount of 40K and 40Ar. Potassium-40 emits a positron to produce argon-40 and the half-life of 40K is 1.25 * 109 years. What is the age of a rock whose 40Ar/40K ratio is 1.42?

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Welcome back everyone. The age of dinosaur fossils is determined by uranium lead dating where the age of a sample is determined through the amounts of lead 206 and uranium 38 in a sample, uranium to 38 decays through a series of nuclear reactions to yield lead 206. How old is a fossil sample? If the sample has a lead 206 to uranium 2 38 ratio of 0.1891, the half life of uranium 2 38 is 4.468 times 10 to the ninth power years. So we're going to begin by noting down the ratio given between lead 206 and uranium 2 38 equal to 0.1891. And because we see that lead to 06 is in the numerator, we're going to go under the assumption that our amount of lead is equal to that value, 0.1891. And because uranium 2 38 is in the denominator, we would assume that the amount that we have at the end of the decay would be one. And so writing out our reaction, we have uranium 238 which decays to form, led to a six as the prompt states. And we want to make an ICF chart to show the progression of this reaction until completion. So for our initial condition of our reactant uranium 2 38 we don't know what it is initially. So we're going to mark it as end zero, which represents our amount of radioisotope one time is equal to zero. So before decay, then for our product lead 206, it hasn't formed yet initially. So we would have zero as its initial amount For the change we're going to go with what is limiting because we have a lower number here for our assumed amount of lead 206 at the end of the reaction, this is going to be what we have for our change. So for the reactant, we would have minus that amount -0.01891. And for our product, we would have plus that amount. Then for the final conditions of the reaction, as we stated, we're assuming we have a value of one for uranium 2 38 and a value of 0.18914 lead 206. Now we need to figure out what end zero is for our initial amount of uranium 2 38. So we would write out the equation which we have technically in our chart already which states that end zero minus 0.1891 Is equal to 1.0. And so solving for N0, we would add plus zero 01891 to both sides Where we would find that our value for the initial amount of our uranium 238 is equal to 1.01891. Now, with this end zero amount, we want to recall that what our reactions describing is radioactive decay. And recall that whenever we have radioactive decay, the reaction follows first order kinetics. And so we want to recall the associated rate law where we would take the natural log of N over N zero set equal to negative one times R D K constant K, which is multiplied by the time that passes T. And so recall that for our decay constant, okay, we can calculate it by taking 0.6 divided by our half life for our radio isotope. And so with this understanding, we can also say that our half life of a radioisotope is calculated by taking 930.6 93 divided by K R decay constant. So now let's go back to our rate law and plug in the values that we know we have N over N zero. So let's go ahead and actually fill in for N for N. We know that that represents our amount of radio isotope after decay occurs. And according to our final conditions of our reaction for uranium 2 38 that would be the value one. So this is our value for N and so filling that in, we have 1.0. And then for N zero, we calculated that to be In our denominator, 1.01891, This is set equal to negative one times our calculation for the decay constant, which as we stated is 0.6 93 where will then take the time that passes in our decay divided by our half life T one half. And so we want to simplify this as much as possible. So for the left hand side, when we take the Ln of our quotient, we get a value of negative 0.187334 set equal to the right hand side which results in the product being negative 0.6 T divided by our half life given in the prompt as 4.468 times 10 to the ninth power years for uranium to 38. So continuing to simplify everything, we now are going to take the result of the quotient on the right hand side. So the left hand side of our equation is the same for the right hand side, we get a result of negative 1.55103 times 10 to the negative 10th power years times T time that passes. And so we need to isolate for time to determine the age of our fossil samples. So we're going to divide by negative 1.55103 times 10 to the negative 10th power years on both sides of our equation. And what we'll find for our results because this will cancel here T is now isolated. So T is equal to a value of 1.2078 times 10 to the eighth power years, which we can say is equal to about 120,000 or 120 million, 800,000 years. But recognize that for our minimum number of sig figs based on our amount of our radio isotope, we have two decimal places in our variable end and then we have five decimal places in our variable and zero. So we want our final answer to be the number of sick fix that matches the smallest number of decimal places in our amount of our radio isotope. And because we see that this is too sick fix. We want our final answer or sorry two decimal places, which is the minimum number of decimal places. We want our final answer to be too sick fix. And so we would round this to two sig figs As 120,000, sorry 120 million years. And this would be our final answer for the amounts or for the age of our fossil sample and this will correspond to choice d in the multiple choice. I hope everything I went through was clear. If you have any questions, leave them down below and I'll see everyone in the next video.

The age of an igneous rock that has solidified from magma can be found by analyzing the amount of 40K and 40Ar. Potassium-40 emits a positron to produce argon-40 and the half-life of 40K is 1.25 * 109 years. What is the age of a rock whose 40Ar/40K ratio is 1.42?

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Key Concepts

Radioactive Decay

Half-Life

Isotope Ratio Dating