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Homework Help: Absorbed Dose of Radiation

  1. Dec 9, 2007 #1
    1. The problem statement, all variables and given/known data

    Suppose you ingest a microgram radioactive material with an initial activity of 10^5/sec and a physical lifetime of 4 hours. Suppose also that all of the energy of each emission of alpha particles with energy of 0.3MeV per particle is deposited throughout your tissue. If you eliminate the material, making a body lifetime of 12 hours, what will be the total dose given to your body, of mass 50kg, after a month?

    2. Relevant equations

    According to my professor,

    Activity = (No/T)exp(-t/T) where T is the lifetime of the substance

    With both a physical half-life, tp, and a body or metabolic half-life, tm,
    N = No(1/2)^(t(1/tp + 1/tm))
    Dose = J/kg

    1 eV = 1.602 x 10^-19 J

    3. The attempt at a solution

    10^5 dis/s X 2419200s (in a month) X (4hrs/12hrs) X (4.806 x 10^-14J)/(dis) X (1/50kg)
    => Dose = 7.75 x 10^-5 J/kg

    Please respond ASAP - the homework is due tomorrow!!! I'll appreciate any help........


  2. jcsd
  3. Dec 10, 2007 #2


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    Homework Helper

    I'm just finding this problem now, so I suppose it may be too late. It seems to me that an integration is required, since you need to know the total number of decays that are released within your body in the course of the month. The combination of radioactive decay and physical elimination gives a lifetime of

    1/(LT) = 1/4 + 1/12 = 1/3 ,

    so the effective lifetime is 3 hours. The total number of decays within your body would then be

    (10^5)·[exp -(t/3)] per second, with t in hours, or

    (3.6·10^8)·[exp -(t/3)] per hour. Integrating this over the 720 hours in a month would give

    -(1/[1/3])·(3.6·10^8)·[exp -(t/3)] , evaluated from t = 0 to t = 720 hours, or

    -(1.08·10^9)·[exp -(720/3)] - { -(1.08·10^9)·[exp -(0/3)] } , which is essentially

    (1.08·10^9) · 1 , since exp -(720/3) is really tiny.

    The isotope is pretty much "gone" within the first 36 hours, since exp(-12) is about 6·10^-6 , so the activity would be less than 1/sec by then.

    With an exposure of 1.08·10^9 decays, the energy release is (1.08·10^9)·(4.806 x 10^-14 J) = 5.19·10^-5 J , or a dosage of 5.19·10^-5 J / 50 kg = 1.04·10^-6 J/kg .

    We're in the same neighborhood, so I suspect you are given an approximate formula for making this calculation. What did they say the answer was?
  4. Dec 12, 2007 #3
    Alas we haven't been provided with a solution to the problem yet.........your procedure seems more concise than mine so I hope that this is indeed the solution to this problem. Thanks for your help!
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