Potassium-40 decays to argon-40 with a half-life of 1.27 * 109 yr. What is the age of a rock in which the mass ratio of 40Ar to 40K is 4.2?

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Hello everyone. So in this video we're dealing with radioactive and nuclear decay of isotopes. So whenever we have this happen we need to go ahead and follow the first order of kinetics. And the integrated rate law for any first order reaction is going to be as follows. So we have the natural log of the concentration at the time equaling to the negative K. Which the decay constant times our time plus the initial concentration. Alright, so let's also recall that the half life is a time that's needed for the amount of a reactant to decrease by 50% of one half life. The half life of a first order reaction is given to us by this equation here to have tea to the half going to natural log of two divided by Kate. So first things first, when you go ahead and calculate the decay constant using the given half life of 5730 hours. So right over here. So we can do is rearrange this equation here and go ahead and isolate the denominator on the right. So the K. R K. Is going to be equal to the natural log of two over T. F. Okay, so put it in my numerical values. Then we have the natural log of two over 5730 years. Alright, so putting everything into my calculator, we have 1.2097 times 10 to the negative four years since we have no concentration that's given to us, we go ahead and actually use the given percentage instead of our concentration. So the information that we're given is going to sort the information out. Given that our concentration at the time Is equal to 1. were also given the initial concentration. Alright so then our constant which we just offer is 1.2097 times to the negative four and T. Is what we don't know and are trying to software so scroll down a little bit. We need to go ahead and solve for our time. So it's going to equal to the natural log of 1.15 equaling two negative 1.2097 times 10 to the negative four multiplied bar T. Let me see here. So again we're multiplying this by T. And then adding the natural log of 18.5. So sort of clearing this up and simplifying this we can say 0.1398 is equaling to the negative of 1.2097 times 10 to the negative four. Go hand multiply by T. again and then we're adding the 2.9178. So combining like terms then you get 1.2097 times 10 to negative four years Times T.  aN:aN:000NaN equaling 2 2.9178 -0.1398. Alrighty. So we have we go ahead and actually divide each side now by all of this right here. So we go ahead and calculate and isolate our team. So by doing that, then I'll get the D. T. Is equal to 2.30 times 10 to the four years the game was isolated for what we're trying to sulfur, which is R. T. Between divide both sides by this here and we get this T. Value so that the value is going to give you my final answer for this problem. Hopefully this all helped.

Potassium-40 decays to argon-40 with a half-life of 1.27 * 109 yr. What is the age of a rock in which the mass ratio of 40Ar to 40K is 4.2?

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Key Concepts

Radioactive Decay

Half-Life

Mass Ratio and Age Calculation