Difference between revisions of "Solutions 7"
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− | It is in total 3.58 mol = 2.1562<math>\cdot</math> 10<sup>24</sup> atoms of potassium in a average human where 12% that is 2.5876 <math>\cdot</math> 10<sup>20</sup> is <sup>40</sup>K | + | # It is in total 3.58 mol = 2.1562<math>\cdot</math> 10<sup>24</sup> atoms of potassium in a average human where 12% that is 2.5876<math>\cdot</math>10<sup>20</sup> is <sup>40</sup>K <br> <math>T_{(1/2)}=1.28\cdot10^{9}=4.0283 \cdot 10^{16}\,s</math><br> <math>\lambda= \frac{\ln 2}{T_{(1/2)}} = \frac{ln2}{4.0283\cdot 10^{16}}=1.7206\cdot 10^{-17}\, s^{-1}</math><br> <math>D=\lambda \cdot N= 2.5876\cdot10^{20}\cdot 1.7206\cdot 10^{-17} s^{-1}=4452.45\, Bq</math> |
− | <math> | + | # First calculate the dose per second:<br> <math>4452.45\, Bq \cdot 400\, keV Bq^{-1} = 1.78\cdot 10^{6} keV s^-1</math><br> Then calculate it to joule per second:<br> <math>1.602\cdot 10^{-16}\, J/keV \cdot 1.78\cdot 10^{6}\, keV/s = 2.8534\cdot 10^{-10}\, J/s</math><br> which gives: <br> <math>2.8534\, J/s \cdot\frac{10^{-10}}{70\, kg} = 4.076\cdot 10^{-12} J/(kg s)</math><br> As the dose can be assumed to be constant trough a human life for each year we will recive a dose of:<br><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> |
− | <math>\lambda= \frac{\ln 2} | ||
− | <math>D=\lambda \cdot N= 2.5876\cdot10^{20}\cdot 1.7206\cdot 10^{-17} s^{-1}=4452.45\, Bq</math> |
Revision as of 12:09, 22 June 2012
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:
- It is in total 3.58 mol = 2.1562
1024 atoms of potassium in a average human where 12% that is 2.5876 1020 is 40K - First calculate the dose per second:
Then calculate it to joule per second:
which gives:
As the dose can be assumed to be constant trough a human life for each year we will recive a dose of: