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Nuclear equation of the decay, decay constant, # of atoms

  1. Oct 27, 2016 #1
    1. The problem statement, all variables and given/known data
    A sample of iodine contains 1 atom of the radioactive isotope iodine 131 (131I) for every 5 * 107 atoms of the stable isotope iodine-127. Iodine has a proton number of 52 and the radioactive isotope decays into xenon 131 (131Xe) with the emission of a single negatively charged particle.

    (a) What particle is emitted when iodine 131 decays? Write the nuclear equation which represents the decay.

    (b) The diagram shows how the activity of a freshly prepared sample of the iodine varies with time. Use the graph to determine the decay constant of iodine 131. Give your answer in s-1.

    f850032ffa6a.jpg

    (c) Determine the number of iodine atoms in the original sample.

    Answers: (b) 1.0 * 10-6 s-1, (c) 1.0 * 1018

    2. The attempt at a solution
    (a) Is it something like: 13152I → 12752I → 131?Xe?

    (b) N = N0 e-λ t, where the decay consant is λ, N = remaining number of atoms, N0 = original number of atoms, t = time.

    We have N0 = 5 * 107 atoms, how do we find N?
     
  2. jcsd
  3. Oct 27, 2016 #2

    jbriggs444

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    Science Advisor

    You do not have that N0 = 5 x 107 atoms. You are asking how to find N0.

    The graph has some useful information. It is expressed in units of Bequerels. If you are a dummy like myself you could Google that unit.

    If you have a rate of decay in decays per second and you have already calculated a rate of decay in decays per nucleon per second, the next step seems obvious.
     
  4. Oct 27, 2016 #3
    So: ln N / N0 = - λ t. I take N = 0.5 * 104 Bq, N0 = 5 * 107 atoms, t = 16 days = 1 382 400 s.

    So: ln 0.5 * 104 / 5 * 107 = - λ * 1 382 400 → λ = 6.66 * 10-6 s-1.
     
  5. Oct 27, 2016 #4
    an important relationship you should know....text book stuff is that
    decay constant λ = (ln2)/ t1/2 ... t1/2 = half life
    or, if you prefer it t1/2 = (ln2)/λ
     
  6. Oct 27, 2016 #5
    But we don't know half-life.
     
  7. Oct 27, 2016 #6
    you have a graph !!!
     
  8. Oct 27, 2016 #7
    It shows activity per time, not half-life.
     
  9. Oct 27, 2016 #8
    the activity is decreasing with time....do you understand what 'half life' means in radioactive decay ?
     
  10. Oct 27, 2016 #9
    The time required to lower the activity of the sample to half of the original value?
     
  11. Oct 27, 2016 #10
  12. Oct 27, 2016 #11
    Activity 1 * 104 Bq with time 8 days → = 691 200 s.

    λ = ln 2 / 691 200 = 1 * 10-6 s-1. Should be correct?
     
  13. Oct 27, 2016 #12
    ideally you should take a few values of half life and get an average
    you have gone from 2 to 1 x 104Bq
    you could go ,from 1 to 0.5 0r any other combination....as long as it is 1/2
    you have changed days into seconds...this is good !
     
  14. Oct 28, 2016 #13
    Could you suggest what to do in (a)?
    Is it something like: 13152I → 12752I → 131?Xe?

    I know α, β and γ particles, is this the situation when they are applicable? Like 13152I → 12750I + 42α? But how do I approach 131?Xe then?

    And if it's not these particles, what should be done?
     
  15. Oct 28, 2016 #14
    131
    It is the isotope Iodine131 that decays to Xenon131 so the nucleus contains the same number of nucleons ( protons + neutrons)
    A negative particle was emitted....any ideas what the negative particle is and how this changes atomic number and nucleon number (mass number) of the nucleus?
     
  16. Oct 28, 2016 #15
    So it is: 13152I → 13153Xe + 0-1β. Correct?

    But why do they say something about
    some "isotope iodine-127" and etc. I thought that the 131 sample decays to 127 sample and then decays to 131 Xe.

    And also what to do in (c)?
    N = N0 e-λ t, where the decay consant is λ, N = remaining number of atoms, N0 = original number of atoms, t = time. We need to find N0, but what is the current number of atoms N?

    I also tried to use dN / dt = - λ N to find N, where dN / dt = 1 * 104 Bq activity and λ = 1 * 10-6 s-1, but that gives 1010, not 1018 as in the answer.
     
    Last edited: Oct 28, 2016
  17. Oct 28, 2016 #16
    You have it correct. The information is needed for part (c)
    They want the total number of iodine atoms...not just the radioactive ones. They have told you the proportion of atoms that are radioactive
     
  18. Oct 28, 2016 #17

    gneill

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    Staff: Mentor

    Just a note on the question accuracy -
    Looks like a typo in the question. According to my periodic table Iodine has a proton number of 53, not 52. Xenon has a proton number of 54. So the transition from Iodine to Xenon involves the number of protons going from 53 to 54.

    This revelation won't change the type of particle emitted, but it will make your transition equation more realistic.

    Just bump the proton numbers up by one on the atoms.
     
  19. Oct 28, 2016 #18
    So, for (c):
    We have 1 atom of radioactive iodine 131 for every 5 * 107 atoms of stable isotope. But it's just one atom, we don't know how much atoms are there in total.

    In N = N0 e-λ t -- how to find N so it would be possible to calculate N0?
     
  20. Oct 28, 2016 #19

    gneill

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    Staff: Mentor

    Your approach using the derivative in post #15 was fine. You found the number of radioactive atoms in the original sample. Each of those atoms will be accompanied by 5 * 107 atoms of stable isotope.
     
  21. Oct 28, 2016 #20
    But 1010 * 5 * 107 = 5 * 1017, not 1 * 1018.
     
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