Difference between revisions of "Solutions 7"

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Return to [[Problem Solving Sets]]
 
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For these exercises it might be usfull to have wallet cards they can for instance be found at [http://www.nndc.bnl.gov/nudat2/indx_sigma.jsp Nudat 2]
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These exercises it might be usfull to have wallet cards they can for instance be found at [http://www.nndc.bnl.gov/nudat2/indx_sigma.jsp Nudat 2]
  
 
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#<math>\frac{m}{M_{m}}\cdot N_{a}\cdot \% = N  </math> <br> <math>\frac{N}{\lambda}=D</math><br> Where m is mass, M<sub>m</sub> is molar mass, N<sub>a</sub> is avogadros constant,&nbsp;% is the percentage N is the number of atoms <span class="texhtml">λ</span> disintigration constant.
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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
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#Since the half-life of potassium is so low we can assume that the amount lost due to decay is negligible. The amount of potassium is constant<br> <math>G=\frac{D\cdot t\cdot E}{m}</math> <br>where G is dose in J/kg, D is the Disintegration rate, t is the time and E is the energy deposited by the radiation.
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<math>T(1/2)=1.28\cdot10^{9}=4.0283 10^{16}\,s</math>
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<math>\lambda= \frac{\ln 2}/{T_{(1/2)}} = \frac{ln2}/{4.0283 10^16}=1.7206\cdot 10^{-17} s^{-1}</math>
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<math>D=\lambda \cdot N= 2.5876\cdot10^{20}\cdot 1.7206\cdot 10^{-17} s^{-1}=4452.45\, Bq</math>
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#<math>D\cdot\lambda=N</math><br><math>\frac{N}{N_{a}}\cdot M_{m}=m</math><br>
 
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#First calculate the effective half-life in the body <math>T_{eff}=\frac{T_{Bi}\cdot T_{phy}}{T_{Bi}+T_{phy}}</math>
 

Revision as of 11:52, 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

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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 10^{16}\,s[/math] [math]\lambda= \frac{\ln 2}/{T_{(1/2)}} = \frac{ln2}/{4.0283 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]

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