K-ar age dating

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K-Ar dating calculation - Life on earth and in the universe - Cosmology & Astronomy - Khan Academy

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  • age - How to sample for K-Ar dating? - Earth Science Stack Exchange.
  • Your Answer.
  • Potassium-Argon Dating!
  • Potassium-argon (K-Ar) dating (video) | Khan Academy.
  • Democracy, literally, rule by the people. Atom, smallest unit into which matter can be divided without the release of electrically charged particles. The sample amount depends on the age because you need enough signal strength to measure the radiogenic argon component precisely.

    Potassium-argon dating | www.hiphopenation.com

    Having the incorrect amount is akin to trying to measure micrometers with a yardstick. For K measurements 30 mg in our lab is routine. For the argon measurement we usually use about 2 mg; for very young rocks up to mg for rocks with ca. We can date volcanic rocks of less than years using this technique and have dated historic eruptions successfully in our lab. Usually customers can say if the rocks are very young, so I usually prepare the sample amount accordingly.

    I had a couple of completely blind samples where I had no idea of the age of the sample a priori, I ran a test with a ca 4 mg which yielded almost no gas, so I re-ran with more material. Regarding sampling technique itself- look for minerals with more potassium, such as muscovite, biotite, amphiboles, or more felsic rocks. If the rocks are obviously weathered, they are not suitable. This is going to be equal to-- and I won't do any of the math-- so we have 1 milligram we have left is equal to 1 milligram-- which is what we found-- plus 0. And then, all of that times e to the negative kt.

    And what you see here is, when we want to solve for t-- assuming we know k, and we do know k now-- that really, the absolute amount doesn't matter. What actually matters is the ratio.

    Because if we're solving for t, you want to divide both sides of this equation by this quantity right over here. So you get this side-- the left-hand side-- divide both sides. You get 1 milligram over this quantity-- I'll write it in blue-- over this quantity is going to be 1 plus-- I'm just going to assume, actually, that the units here are milligrams. So you get 1 over this quantity, which is 1 plus 0.

    That is equal to e to the negative kt. And then, if you want to solve for t, you want to take the natural log of both sides. This is equal right over here. You want to take the natural log of both sides. So you get the natural log of 1 over 1 plus 0. And then, to solve for t, you divide both sides by negative k. So I'll write it over here. And you can see, this a little bit cumbersome mathematically, but we're getting to the answer.


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    • So we got the natural log of 1 over 1 plus 0. Well, what is negative k? We're just dividing both sides of this equation by negative k. Negative k is the negative of this over the negative natural log of 2 over 1. And now, we can get our calculator out and just solve for what this time is. And it's going to be in years because that's how we figured out this constant.

      So let's get my handy TI First, I'll do this part. So this is 1 divided by 1 plus 0. So that's this part right over here. That gives us that number. And then, we want to take the natural log of that. So let's take the natural log of our previous answer. So it's the natural log of 0.

      K–Ar dating

      It gives us negative 0. So that's this numerator over here. And we're going to divide that. So this number is our numerator right over here. We're going to divide that by the negative-- I'll use parentheses carefully-- the negative natural log of that's that there-- divided by 1.

      So it's negative natural log of 2 divided by 1. E9 means times 10 to the ninth. And I closed both parentheses. And now, we need our drum roll.