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Cobalt decay and activity

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  1. Sep 1, 2015 #1

    Rectifier

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    The problem statement
    The Cobalt-60 isotope is often used in medicine. It has a half-life of 5.25 years.
    a) What is the activity of 1μg Cobalt in Bq (decays/second)?
    b) You find a sample of Cobalt that has a mass of 10 µg and has an activity of 5.0MBq. How old is that sample?

    This problem was translated from Swedish.

    The attempt at a solution
    a)
    Activity is given by:
    ## A_{Bq}= n N_A \frac{ln(2)}{t_{\frac{1}{2}}} ##
    where ## t_{\frac{1}{2}} = 1.656 \cdot 10^8 ## since ## 5.25 years =1.656 \cdot 10^8 seconds ## and ##N_A## is avogados constant which is ## 6.022×10^{23} mol^{−1} ## but what is n?

    I guess that that is the amount of moles in 1μg Cobalt. Wolfram tells me its
    ## 1.6968366×10^{-8} mol ##

    Thus the activity is
    ## A_{Bq}= n N_A \frac{ln(2)}{t_{\frac{1}{2}}} = 1.6968366 \cdot 10^{-8}\cdot 6.022\cdot10^{23}\cdot \frac{ln(2)}{1.656 \cdot 10^8}=
    4.27707 \cdot 10^7##

    b)
    ## A(t) = A_0 0.5^{\frac{t}{ t_{\frac{1}{2}} }} ##
    where
    A(t) = 5.0MBq
    A(0) is the activity of 1μg Cobalt
    ##t_{\frac{1}{2}}## = ## 1.656 \cdot 10^8 ##

    ## A(t) = A_0 0.5^{\frac{t}{ t_{\frac{1}{2}} }} \\ 5.0 \cdot 10^6 = 4.27707 \cdot 10^7 \cdot 0.5^{\frac{t}{ 1.656 \cdot 10^8} } \\ t =5.12801×10^8 s ##
    Which is about 16.26 years.

    Is that solution correct?
     
    Last edited: Sep 1, 2015
  2. jcsd
  3. Sep 1, 2015 #2
    The activity of 1 gram of Cobalt-60 is 44 TBq, so 44 trillion Bq. The activity of a μg (microgram) of Cobalt-60 would be 44,000,000,000,000 divided by 1,000, which is 44,000,000,000 (44 billion Bq). So the answer to part a): The activity of 1 microgram of Cobalt-60 is 44 billion (44,000,000,000) Bq.

    I'm afraid I can't help with part b) though. Hope this helped.
     
  4. Sep 1, 2015 #3

    mfb

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    Approximately, but looking that up is certainly not the intended solution if the half-life is given.
    Also note that a gram has 1 million microgram, not 1000.

    @Rectifier: Correct.
     
  5. Sep 1, 2015 #4

    Rectifier

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    Thank you!!!!!!!
     
  6. Sep 1, 2015 #5

    Bystander

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    Works.
     
  7. Sep 1, 2015 #6
    Apologies, thanks for the correction.
     
  8. Sep 1, 2015 #7
    For part a), did you find the number of moles for Co-60 or Co-59 (natural cobalt)?

    For part b), should A0 be for 1 ug or 10 ug?
     
  9. Sep 1, 2015 #8

    mfb

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    Good points. I should have checked the numbers more carefully.
     
  10. Sep 2, 2015 #9

    Rectifier

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    a) that was the natural cobalt its not specified in the problem :(
    b) ah darn, I guess I have to recalculate A_0 for 10 micro-grams. Is it just 10 times bigger by any chance?
     
  11. Sep 2, 2015 #10
    Bingo!
     
  12. Sep 2, 2015 #11

    Rectifier

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    I am not entierly sure about what to do with a) thoug? Any tips?
     
  13. Sep 2, 2015 #12
    Just correct this using the result for M from your other thread.
     
  14. Sep 2, 2015 #13

    mfb

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    It is, you have pure Co-60.
     
  15. Sep 2, 2015 #14

    Rectifier

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    Attempt 2 :D

    a)

    ## A_{Bq}= n N_A \frac{ln(2)}{t_{\frac{1}{2}}} ##
    ## t_{\frac{1}{2}} = 1.656 \cdot 10^8 ##
    ##N_A = 6.022 \cdot 10^{23} mol^{−1} ##
    ## n = \frac{n}{M} = \frac{10^{-6}}{59.933817059} = 1.668507 \cdot 10^{-8} ##

    ## A_{Bq}= n N_A \frac{ln(2)}{t_{\frac{1}{2}}} = 1.668507 \cdot 10^{-8} \cdot 6.022 \cdot 10^{23} \cdot \frac{ln(2)}{1.656 \cdot 10^8}= 4.20566 \cdot 10^7##


    b)
    ## A(t) = A_0 0.5^{\frac{t}{ t_{\frac{1}{2}} }} ##
    where
    A(t) = 5.0MBq
    ##t_{\frac{1}{2}} = 1.656 \cdot 10^8 ##

    A(0) is the activity of 10μg Cobalt
    ## n = \frac{n}{M} = \frac{10 \cdot 10^{-6}}{59.933817059} = 1.6685 \cdot 10^{-7}## (turns out activity of 1μg * 10 = activity of 10μg)

    ## A_{Bq} = n N_A \frac{ln(2)}{t_{\frac{1}{2}}} \\ A_{Bq} = 1.6685 \cdot 10^{-7} 6.022 \cdot 10^{23} \frac{ln(2)}{1.656 \cdot 10^8} = \\ A_{Bq} = 4.20564 \cdot 10^8##

    ## A(t) = A_0 0.5^{\frac{t}{ t_{\frac{1}{2}} }} \\ 5.0 \cdot 10^6 = 4.20564 \cdot 10^8 \cdot 0.5^{\frac{t}{ 1.656 \cdot 10^8 }} = 1.05889 \cdot 10^9 s ##
    Or also 33.58 years

    Is this right then? :)
     
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