One Response to “For 2.0 mole of the radioactive isotope molybdenum-99, half-life = 67 hours, how many disintegrations?”

1. cattbarf Says:
   August 30th, 2008 at 1:33 am

   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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