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When Po decays, the product is Pb. The half-life of this decay process is 1.78 ms. If the initial sample contains $3.4\times 10^{17}$ parent nuclei, how many are remaining after 35 ms have elapsed? What kind of decay process is this (alpha, beta, or gamma)?
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Final Answer
This is alpha decay. $4.1 \times 10^{11}$ nuclei remain.
Solution Video

OpenStax College Physics Solution, Chapter 31, Problem 11 (Test Prep for AP® Courses) (1:18)

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Video Transcript

This is College Physics Answers with Shaun Dychko. Pulonium decays into lead and the first thing we'll answer is what type of decay process is this? And it's a process that involves changing an atomic number from 84 to 82; pulonium has atomic number 84 and lead has 82 protons and that process that does that is <i>α</i>-decay and also two neutrons are packaged along with those two protons to make a total of four nucleons bound together into this <i>α</i>-particle or helium nucleus. Now, the number of the parent nuclei that will remain after 35 milliseconds is given by this formula here; this is the number of parent nuclei remaining, given an initial number which in this case is 3.4 times 10 to the 17 times <i>e</i> to the negative decay constant times time. This decay constant is natural log of 2 over the half-life so we substitute that in and we calculate that 3.4 times 10 to the 7— initial number of parent nuclei— times <i>e</i> to the negative of the log of 2 divided by 1.78 milliseconds—half-life— times 35 milliseconds means there will be 4.1 times 10 to the 11 nuclei remaining after 35 milliseconds.