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<p class="MsoNormal"><span style="font-family:"Arial",sans-serif">Someone asked about SA Exercise 21.11. It asks the following:<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif">“It takes 45 hours for a 6.00 mg sample of sodium-24 to decay to<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif">0.750 mg. What is the half-life of sodium-24?”<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif">There’s a couple of ways to do this:<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif">1) You can use the 1<sup>st</sup>-order integrated rate eqn to find k (the decay constant,<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif">i.e. the rate constant) and then use the half-life eqn to get t_1/2.<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif">ln([N]t/[No]) = -k*t<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif">k = - ln([N]t/[No])/t = - ln(0.750/6.00)/45 hr = 4.621 x 10^-2 hr^-1<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif">t_1/2 = 0.693/(4.621 x 10^-2 hr^-1) = 14.996 hr = 15 hr<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif">2) This way is maybe a little easier but harder to see right off.<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif">Take the ratio of the final mass to initial mass:<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif">0.750/6.00 = 0.125<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif">That’s 12.5%. That’s one of the “magic” numbers. To get to 12.5% it<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif">takes 3 half-lives,<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif">100% </span><span style="font-family:Wingdings">à</span><span style="font-family:"Arial",sans-serif"> 50%
</span><span style="font-family:Wingdings">à</span><span style="font-family:"Arial",sans-serif"> 25%
</span><span style="font-family:Wingdings">à</span><span style="font-family:"Arial",sans-serif"> 12.5%
</span><span style="font-family:Wingdings">à</span><span style="font-family:"Arial",sans-serif"> 6.25%
</span><span style="font-family:Wingdings">à</span><span style="font-family:"Arial",sans-serif"> 3.125% etc.<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif">Thus, since the ratio of the masses (concentrations) is 0.125 (12.5%)<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif">that signifies 3 half-lives.<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif">The half-life for 1<sup>st</sup>-order processes is constant. That means if the<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif">total time is 45 hours and that represents 3 half-lives the half-life<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif">is 15 hours.<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif">Either way that’s not one of the possible answers. I’ve reported<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif">it to Pearson.<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif">Dr. Zellmer<o:p></o:p></span></p>
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