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Radioactive Decay and Activity

  1. Jun 10, 2008 #1
    Hi there,

    I was wondering if anyone could help me with this question ...

    "A device used in radiation therapy for cancer contains 0.50g of cobalt (60-Co-27). The half life of cobolt is 5.27 years. Determine the activity of the radioactive material."

    Working out:
    (To find the decay constant)
    T1/2= ln2/[lamba]
    [lamba]= ln2/5.27 years
    =0.1315

    (To find activity)
    dN/dt= -[lamba]N
    ...

    How do you calculate for N?! (I've been trying to use the Avogadro's number but keep on obtaining really bizarre numbers ... Perhaps I'm not using it correctly)
    N=0.0005kg / 60u (60 x 1.6605x10^27) x Avagadro's Number

    The correct answer is: 2.1x10^13 Bq

    Any help will be much appreciated
     
  2. jcsd
  3. Jun 10, 2008 #2

    Kurdt

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    You only need avagadro's number if you're working in moles. Since you're not the mass divided by the mass/atom will give you the number of atoms.
     
  4. Jun 10, 2008 #3

    Astronuc

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    Work out the units. As Kurdt mentioned, one knows the mass, so using the number of atoms/mole and atomic mass one can find N.

    Then the decay constant must be in the correct units. One shows [lambda] in yrs-1, but it is usually in s-1.
     
  5. Jun 10, 2008 #4
    Ohh! I see ... Thank you so much! :)
     
  6. Jun 10, 2008 #5

    Kurdt

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    It will definitely need to be in seconds if you want to find the activity.
     
  7. Jun 10, 2008 #6
    I just came across another question that I'm stuck with ... I'm new to this forum, and so I hope its ok for me to post this question here?

    "A beam of nuclei is used for cancer therapy. Each nucleus has an energy of 140 MeV, and the relative biological effectiveness (RBE) of this type of radiation is 16. The beam is directed onto a 0.17-kg tumor, which receives a biologically equivalent dose of 1.3 Sv. How many nuclei are in the beam?"

    Absorbed Dose (AD) = Energy absorbed (Q)/Mass of absorbed Material
    So... I converted 140MeV =>2.24x10^-11J

    Biologically Equivalent Dose (BED) = (AD) x (RBE)
    (AD) = BED/RBE
    =>1.3/16 = 0.08125 Gy

    And I was stuck here ...
     
  8. Jun 10, 2008 #7
    The correct answer is: 6.2x10^8
     
  9. Jun 10, 2008 #8

    Kurdt

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    Since AD is energy per unit mass and one knows the mass of the tumour and the energy per nuclei can you see how to work it out from there?
     
  10. Jun 10, 2008 #9
    0.08125 = 2.24x10^-11J / 0.17kg
    => 0.08125/1.20545x10^-18
    =6.16x10^8 (3sf)

    Thank you Kurdt!!
     
  11. Jun 10, 2008 #10

    Kurdt

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    No problem. Remember that dimensional analysis can be very useful so always pay attention to the units of the things you have and the things you want to find.
     
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