I've got some A2 physics homework and unfortunately i can't seem to do the first question :/

The question is: 61Co has a half life of 100 minutes. how long will it take for the radioactive isotope to reduce to 80% of the initial value?

The answer is 32 minutes, but of course i'll need the workings before it to actually prove i can do the question.

Thank you for any help given! (if any is anyway).

2 Answers

  • 8 years ago
    Favourite answer

    The fraction remaining after t minutes is:


    So if 80% remains we need to solve:


    which we do by taking logs (the base is unimportant as long as we use the sane base throughout):

    log(0.8)=(-t/100) log(2)


    t=- [log(0.8)/log(2)] * 100


    Source(s): Also posted on MathHelpBoards with better notation at:
  • 4 years ago

    carbon 14s 0.5 existence is 5760 years/ln 2 = 8309.9234 years = tau, residing tissue incorporates 1 million x10 to the -12 for each gram of carbon. 8309.9234 ln 8/a million = 17279.95127 years previous. tau is a a million/2 existence time consistent it truly is from the answer to a differential equation m = mo exp (-t/tau) that you should understand a thanks to have the flexibility to remedy. the first answer became sturdy adequate searching on how precise you want to be.

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