Difference between revisions of "Solutions 7"

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This solution set contains a large number of errors, and is under revision. It will be updated as soon as possible.

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These exercises it might be usfull to have wallet cards they can for instance be found at Nudat 2

1:

1. It is in total 3.58 mol = 2.1562[math]\cdot[/math] 10²⁴ atoms of potassium in a average human where 12% that 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 trough a human life for each year we will recive 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 renderer 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]
   for a person of 70 kg this gives a dose of 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]
   Then the total dose will be:
   [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]