Comments on: For 2.0 mole of the radioactive isotope molybdenum-99, half-life = 67 hours, how many disintegrations? http://molybdenuminfo.com/2008/08/27/for-20-mole-of-the-radioactive-isotope-molybdenum-99-half-life-67-hours-how-many-disintegrations/ Answering your molybdenum questions Sat, 11 Sep 2010 00:55:41 +0000 http://wordpress.org/?v=2.6.3 By: cattbarf http://molybdenuminfo.com/2008/08/27/for-20-mole-of-the-radioactive-isotope-molybdenum-99-half-life-67-hours-how-many-disintegrations/#comment-71 cattbarf Sat, 30 Aug 2008 01:33:27 +0000 http://molybdenuminfo.com/?p=86#comment-71 Basically, 2 moles = 1.2x10^24 atoms. We want to know how many have gone after 17 minutes. Half-life is really not a good way to go after this. I favor using the decay constant, which is more flexible. This constant, k is obtained from the half-life. 67(60) k = 0.693 (we want an answer based on minutes) k = 0.693/4020=1.72x10-4 min appx. now C= Co exp(-kt) or exp is 2.7182 So C(17 min)= 1.2x10^24exp(-17x1.72x10-4) C(17 min)=1.2x10-24exp(-.0029) and solve. The difference between 1.2x10-24 and C(17 min) is what you need. Basically, 2 moles = 1.2×10^24 atoms. We want to know how many have gone after 17 minutes.

Half-life is really not a good way to go after this. I favor using the decay constant, which is more flexible. This constant, k is obtained from the half-life. 67(60) k = 0.693
(we want an answer based on minutes)
k = 0.693/4020=1.72×10-4 min appx. now C= Co exp(-kt) or exp is 2.7182
So C(17 min)= 1.2×10^24exp(-17×1.72×10-4)
C(17 min)=1.2×10-24exp(-.0029) and solve. The difference between 1.2×10-24 and C(17 min) is what you need.

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