Solutions 7

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Radiation Dosage and Radiation Protection

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For the following exercises it is recommended to have Nuclear Wallet Cards close at hand. One example can be found here: Nudat 2

1:

  1. It is in total 3.58 mol = 2.1562[math]\cdot[/math] 1024 atoms of potassium in an average human, 12% of this is  2.5876[math]\cdot[/math]1020 is 40K
    [math]T_{(1/2)}=1.28\cdot10^{9}=4.0283 \cdot 10^{16}\,s[/math]
    [math]\lambda= \frac{\ln 2}{T_{(1/2)}} = \frac{ln2}{4.0283\cdot 10^{16}}=1.7206\cdot 10^{-17}\, s^{-1}[/math]
    [math]D=\lambda \cdot N= 2.5876\cdot10^{20}\cdot 1.7206\cdot 10^{-17} s^{-1}=4452.45\, Bq[/math]
  2. First calculate the dose per second:
    [math]4452.45\, Bq \cdot 400\, keV Bq^{-1} = 1.78\cdot 10^{6} keV s^-1[/math]
    Then calculate it to joule per second:
    [math]1.602\cdot 10^{-16}\, J/keV \cdot 1.78\cdot 10^{6}\, keV/s = 2.8534\cdot 10^{-10}\, J/s[/math]
    which gives:
    [math]2.8534\, J/s \cdot\frac{10^{-10}}{70\, kg} = 4.076\cdot 10^{-12} J/(kg s)[/math]
    As the dose can be assumed to be constant throughout a human life, each year we will receive a dose of:
    [math]3.145\cdot 10^7 \, \frac{s}{year} \cdot 4.076\cdot 10^{-12}\, \frac{J}{kg\, s} = 1.28 10^{-4}\, \frac{Gy}{year} = 0.13\, \frac{mGy}{year}[/math]

2:

  1. [math]\lambda = 7.3466 \cdot ^{-10}s^{-1}[/math]
    [math]D=\lambda \cdot N \rightarrow N=\frac{D}{\lambda}=\frac{6000 \, Bq}{\lambda}=8.17\cdot 10^{12}[/math]
    which is 1.36[math]\cdot[/math]10-11 mol of 137Cs
  2. Calculate the initial dose rate from the reindeer meat:
    [math]6000\, Bq \cdot 200 keV\cdot 1.6\cdot 10^{-16} \frac{J}{kev}=1.92\cdot 10^{-10}\frac{J}{s}[/math]
    The dose of a person who weights 70 kg equals to 2.74[math]\cdot[/math]10-12Gy/s.
    Then calculate the real half-life
    [math]T_{rel}=\frac{T_{B}\cdot T_{P}}{T_{B}+T{P}}\rightarrow \frac{0.302\cdot 30}{0.302+30}=108.9\, d[/math]
    This gives:
    [math]D_{tot}=\int^{t=\infty}_{t=0}D_{0}\cdot e^{-\lambda t}dt=-\frac{D_{0}}{\lambda}\left[e^{-\lambda t}\right]^{t=\infty}_{t=0}=\frac{D_{0}}{\lambda}(0-1)[/math]
    [math]=\frac{2.74\cdot 10^{-12}}{7.367\cdot 10^{-8}}=0.580\, mGy/year[/math]

3:
ALI (Annual Limit of Intake) is the amount of activity in a certain material that, when consumed, gives a radiaton dose of 20 mSv. There is ALI-values for both inhalation and consummation of radioactivity material.

Both chemical and physical properties can make a substance dangerous. Decay type, half-life and the energy of the radiation determines the dose received. The chemical form and the chemical properties of the material determine if it is accepted or if it gets enriched in the body. For instance a alpha-emitter can be inhaled and do a substantial amount of damage in the body. The combination of a short physical half-life together with a long biological half-life will deposit the greatest amount of dose in the body. Some example of interest are iodine, which accumulates in the thyroid gland, polonium which accumulates in the bones and radon gas which can be accumulated in the lungs.