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	<title>Comments on: For 1.4 mole of the radioactive isotope molybdenum-99, half-life = 67 hours, how many disintegrations?</title>
	<atom:link href="http://molybdenuminfo.com/2008/07/25/for-14-mole-of-the-radioactive-isotope-molybdenum-99-half-life-67-hours-how-many-disintegrations/feed/" rel="self" type="application/rss+xml" />
	<link>http://molybdenuminfo.com/2008/07/25/for-14-mole-of-the-radioactive-isotope-molybdenum-99-half-life-67-hours-how-many-disintegrations/</link>
	<description>Answering your molybdenum questions</description>
	<pubDate>Thu, 09 Sep 2010 03:30:40 +0000</pubDate>
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		<title>By: hfshaw</title>
		<link>http://molybdenuminfo.com/2008/07/25/for-14-mole-of-the-radioactive-isotope-molybdenum-99-half-life-67-hours-how-many-disintegrations/#comment-58</link>
		<dc:creator>hfshaw</dc:creator>
		<pubDate>Mon, 28 Jul 2008 22:02:31 +0000</pubDate>
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		<description>The radioactive decay equation is:

N(t) = N(0)*exp(-L*t)

where N(0) is the quantity of radioactive material present initially

N(t) is the quantity of radioactive material remaining at time t

L is the radioactive decay constant = ln(2)/halflife

The number of decays that have happened between times t=0 and t=t is then N(0) - N(t), so

# decays = N(0)*(1-exp(-L*t))


In this case, L = ln(2)/67 hrs = 1.724*10^-1 min^-1

N(0) = 1.4 moles 

# decays = 1.4 moles * (1 - exp(-1.724*10^-1 min^-1 * 17 min))

# decays = 4.098*10^-3 mol

Multiplying by Avogadro's number to get the number of actual decays:

# decays = 4.098*10^-3 mol* * 6.022*10^23 atoms/mol

# decays = 2.468*10^21 atoms decayed</description>
		<content:encoded><![CDATA[<p>The radioactive decay equation is:</p>
<p>N(t) = N(0)*exp(-L*t)</p>
<p>where N(0) is the quantity of radioactive material present initially</p>
<p>N(t) is the quantity of radioactive material remaining at time t</p>
<p>L is the radioactive decay constant = ln(2)/halflife</p>
<p>The number of decays that have happened between times t=0 and t=t is then N(0) - N(t), so</p>
<p># decays = N(0)*(1-exp(-L*t))</p>
<p>In this case, L = ln(2)/67 hrs = 1.724*10^-1 min^-1</p>
<p>N(0) = 1.4 moles </p>
<p># decays = 1.4 moles * (1 - exp(-1.724*10^-1 min^-1 * 17 min))</p>
<p># decays = 4.098*10^-3 mol</p>
<p>Multiplying by Avogadro&#8217;s number to get the number of actual decays:</p>
<p># decays = 4.098*10^-3 mol* * 6.022*10^23 atoms/mol</p>
<p># decays = 2.468*10^21 atoms decayed</p>
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