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Alpha emission decay: maximum possible delay

  1. Oct 30, 2016 #1
    1. The problem statement, all variables and given/known data
    A tube containing a isotope of radon, 22286Rn, is to be implanted in a patient. The radon has an initial activity of 1.6 * 104 Bq, a half-life of 4 days and it decays by alpha emission. To provide the correct dose, the tube, containing a freshly prepared sample of the isotope, is to be implated for 8 days.

    (a) (i) What are the proton (atomic) number and the nucleon (mass) number of the daughter nucleus produced by the decay of the radon?
    (a) (ii) State one reason why an alpha emitter is preferred to a beta or gamma emitter for such purposes?

    (b) Determine: (i) the decay constant for radon in s-1, (ii) the initial number of radioactive radon atoms in the tube.

    (c)
    The operation to implant the tube has to be delayed. Ignoring the effects of any daughter products of the decay, determine the maximum delay possible if the patient is to receive the prescribe dose using the source.

    Answers: (b) (i) 2.0 * 10-6 s-1, (ii) 8.0 * 109, (c) 40 hours.

    2. The attempt at a solution
    (a) (i) 22286Rn → 21884Po + 42α.
    (a) (ii) No idea. Maybe α is less harmful for health?

    (b) (i) λ = ln 2 / 345 600 = 2 * 10-6 s-1.
    (b) (ii) dN / dt = - λ N → N = (dN / dt) / λ = 16 000 / 2 * 10-6 = 8 * 109 atoms.

    (c) A = A0 e- λ t = 16 000 e- 2 * 10-6 * 691 200 = 4016 Bq. This should be the activity after 8 days. But how to determine the maximum possible delay? I though of something like 0 = 4016 e- 2 * 10-6 * t and solve for t, but doesn't look like mathematically solvable.

    Any help please on bold problems (a) (ii) and (c)?
     
    Last edited: Oct 30, 2016
  2. jcsd
  3. Oct 30, 2016 #2

    kuruman

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    It depends on what the "such purposes are", e.g. therapy vs. diagnosis. You have not specified that.
    The patient has the implant for 8 days and then it is removed. This means that the patient needs a limited dose. What is that dose? This should give you a clue how to answer (c).
     
  4. Oct 30, 2016 #3
    And in general what is the benefit?

    I though the tube "is to be implanted" = planned to be implanted. But then the operation to "implant the tube" is delayed. So the tube will be later put into the patient.

    Plus if the dose is implanted for 8 days it means that the dose should be in the patient for 8 days, no more no less.
     
  5. Oct 30, 2016 #4

    gneill

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    The "dose" means the total amount of radiation particles delivered. The original plan was for the brand new tube to remain for 8 days to deliver the required dose; The 8 day timeframe was for the tube when it was new. A tube with lesser initial activity would need longer to deliver the same number of particles.
     
  6. Oct 30, 2016 #5

    kuruman

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    Don't forget that the isotope decays continuously so if you wait a while, the implant will take more than 8 days to deliver the same number of alpha particles.
     
  7. Oct 30, 2016 #6
    Still don't understand anything.

    We have something put into a person that was designed to last 8 days. So it's obvious that it should be taken from the person right away after the 8 day period.

    OR, depending on the situation, the material in the person could stay for a little longer, but we need to know the activity number when the material becomes dangerous.

    So the activity will be 4016 Bq on day 8. What activity level is harmful to a person? 2000 Bq? 100 Bq?

    Knowing the harmful activity level we can calculate the maximum time. But we are not given it.

    Other than these two situations, I don't quite see a way.
     
  8. Oct 30, 2016 #7

    gneill

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    The time required to deliver a specified dose doesn't (in this problem) involve danger or harm considerations. Provided you don't give a larger dose than is specified, the amount of time it takes isn't an issue. What is important is the total dose delivered over whatever time it takes.

    Here's a hint for shortcut. The dose is equal to the total number of decays that happen while the source is in the patient. That means some amount of the initial material will have decayed in that time period. Find out how much of the material decays during the original planned treatment, either by a count of atoms or just by the fraction of the original material that decays. That will be the dose that you need to be able to match in the delayed scenario, even if the source has to remain in the patient for a (possibly much) longer time.
     
  9. Oct 30, 2016 #8
    A0 = 16 000 Bq and A8 = 4016 Bq. Decayed over the 8 days = 11 984 Bq. Fraction decayed 0.749.

    ln 0.749 = 2 * 10-6 t
    t = 144 508 s = 40.1 hrs.

    I still don't understand this, even though I got the correct answer.

    We have activity 16 000 Bq at day 1, and activity 4016 Bq on day 8 (75 % of the material has decayed). But what's the connection between the decay percentage and the time when the material should be taken back from the patient. Maybe in some simpler words? Why should it matter what's the percentage of the decayed material?

    And also:
    Also didn't get this part. I read some more information on α, β and γ radiation and got an understanding that α has the least penetration potential (it only penetrates skin). So it's less harmful and therefore better for medical purposes? But again, I'm pretty sure there are different procedures that require different type of radiation penetration.
     
  10. Oct 30, 2016 #9

    gneill

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    Let's take a shortcut. The half life of the material is 4 days. The initial dosing time was 8 days. That's two half-lives. So what fraction of the material "disappears" in that time? 1/2 for the first half-life, then half of what's left in the second half-life. Do if you started with N amount (atoms, mass, however you want to judge the amount), the dose was (1/2 + 1/4)N or (3/4)N. That's your 0.749 (really 0.750) fraction.

    If you want to be able to deliver the same dose at a later time, and supposing you can take as much time as necessary for it all to decay, then you need to start with at least this much material. So the question becomes how long can you wait until the original material drops below 3/4 of what you started with.
    As I've repeatedly stated, the dose is the total number of decays. It's not the activity itself. Lower starting activity just means it will take longer to deliver the same number of decays to the patient (dose).
    You're getting the picture. Usually when a source is implanted it's meant to affect a specific, local target. For example, it might be implanted next to or inside of a particular tumor. Ideally you don't want to irradiate anything other than the target. So a small penetration depth gives better control of the affected area and is a good thing when it can be used. Each radiation type has hazards for side effects according to how it interacts with the tissue molecules and atoms.
     
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