Question
What fraction of the $^{40}\textrm{K}$ that was on Earth when it formed $4.5\times 10^{9}$ years ago is left today?
Question by OpenStax is licensed under CC BY 4.0.
Final Answer
$0.087$
Solution Video

# OpenStax College Physics Solution, Chapter 31, Problem 55 (Problems & Exercises) (2:12)

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This is College Physics Answers with Shaun Dychko. Four and a half billion years ago when the Earth was formed it was composed of a certain amount of potassium-40 isotope and we want to figure out, what fraction of that is remaining today? So we know the half-life of potassium-40 is 1.28 billion years and we'll use this formula number 36 from chapter 31 to solve the question but then I have an alternative method down here using this one-half to the power of <i>t</i> over <i>t</i> one-half here. I'll explain what that is in a second, it's an alternative method. So the number of atoms of a radioactive material is the initial number of atoms times <i>e</i> to the negative decay constant times <i>t</i> and then we can divide these two to get the fraction of the atoms that still remain. And so we have to substitute for lambda; it's logarithm of 2 divided by the half-life; that's equation 37 and then substitute that in for lambda here and so the fraction of potassium-40 remaining is <i>e</i> to the negative logarithm of 2 times <i>t</i> over <i>t</i> one-half. So <i>t</i> one-half for potassium-40 is 1.28 times 10 to the 9 years and the Earth is 4.5 times 10 to the 9 years old and multiply that by logarithm 2 and raise <i>e</i> to the negative of that and we get 0.087. So there is 8.7 percent of the potassium-40 is still remaining in the Earth. An alternative method would be to say that the number of atoms equals the original number of atoms multiplied by one-half to the power of the number of half-lives that have elapsed. And so when you go an amount of time divided by the half-life, this gives you the number of half-lives, the number of times that one should multiply <i>N naught</i> by a half. And so we divide both sides by <i>N naught</i> and we get <i>N</i> over <i>N naught</i> is a half to the power of 4.5 times 10 to the 9 years divided by 1.28 times 10 to the 9 years and this also gives 0.087 as we expect.