One Response to “For 1.4 mole of the radioactive isotope molybdenum-99, half-life = 67 hours, how many disintegrations?”

1. hfshaw Says:
   July 28th, 2008 at 10:02 pm

   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

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