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A question related to radioactivity.

  1. May 4, 2011 #1
    1. A cobalt-60 source having a half-life of 5.27 years is calibrated and found to have an
    activity of 3.50 × 105 Bq. The uncertainty in the calibration is ±2%.
    Calculate the length of time, in days, after the calibration has been made, for the stated
    activity of 3.50 × 105 Bq to have a maximum possible error of 10%.




    A= -лN, where A= Activity, л=Decay Constant and N=Original number of nuclides.

    n=Ne^(-лt), where n=number of undecayed nuclides, N= original number of nuclides, л=decay constant and t=time.




    3. The attempt at a solution: None

    I am utterly stumped at this one. Truthfully I didn't even understand the question properly. I can't make a connection between this uncertainty and decay. I have the solution but that didn't help me much. Here it is:

    source must decay by 8%
    A = A0 exp(–ln2 t / T½) or A/ A0 = 1 / (2t/T)
    0.92 = exp(–ln2 × t / 5.27) or 0.92 = 1 / (2t/5.27)
    t = 0.634 years
    = 230 days
    (allow 2 marks for A/ A0 = 0.08, answer 7010 days
    allow 1 mark for A/ A0 = 0.12, answer 5880 days)


    Could someone please explain to me what exactly the question asks for, and how the quantities are related? Thanks in advance.
     
  2. jcsd
  3. Apr 26, 2012 #2
    CIE a level physics paper 4. the question was made to be unanswered i guess.
     
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