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Cow’s milk produced near nuclear reactors can be tested for as little as 1.00 pCi of $^{131}\textrm{I}$ per liter, to check for possible reactor leakage. What mass of $^{131}\textrm{I}$ has this activity?
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Final Answer
$8.12\times 10^{-18}\textrm{ g}$
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OpenStax College Physics Solution, Chapter 31, Problem 49 (Problems & Exercises) (2:57)

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This is College Physics Answers with Shaun Dychko. We want to know what mass of iodine-131 has an activity of 1 picocurie. So we gather some information here; we know that the molar mass of iodine-131 is 130.906 grams per mol and we found that in appendix A and the half-life of iodine-131 is 8.040 days and we found that in appendix B. So the activity of a radioactive substance is natural logarithm of 2 times the number of atoms divided by the half-life. And the number of atoms we can express in terms of mass and then we'll rearrange and solve this for mass because we know what the activity is and everything else here and the only thing we don't know is mass. So the number of atoms is the mass divided by the molar mass so this is grams divided by grams per mol which makes this ratio have units of mol's and then we multiply by Avogadro's number which is number of atoms per mol and so those mol's cancel leaving us with number of atoms. And so we'll replace <i>n</i> with all of this and that's what I do here. Well actually what I did is I rearranged this one for <i>n</i> and then say equated it to this thing, that's what happened here. So this rearranged for <i>n</i> is multiplied by half-life divided by log 2 on both sides and then you get this part here. So <i>R</i> times half-life over log 2 is the number of atoms and that equals all of this here; mass divided by molar mass times Avogadro's number. So then we solve this for <i>m</i> the mass by multiplying by the molar mass and dividing by Avogardo's number on both sides here. So the mass is the activity times the half-life times the molar mass divided by log 2 times Avogadro's number. So that's 1 picocurie which is times 10 to the minus 12 curies times 3.70 times 10 to the 10 becquerel's per curie because we need to have <i>mks</i> units and becquerel is the <i>mks</i> unit of radioactivity. So put that <i>B</i> back there then multiply by the half-life expressed in seconds so multiplying 8.040 days by 24 hours per day times 3600 seconds per hour and them multiply by the molar mass— 131.906 grams per mol— and divide by log 2 times Avogadro's number and we get 8.12 times 10 to the minus 18 grams of iodine-131 would have an activity of 1 picocurie.