Difference between revisions of "Solutions 2"

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#1000g Th(NO3)4 = 2.083 mol arrow N(Th)= 1.25<math>\cdot</math>10<sup>24</sup> atoms. This is natural thorium, where the equilibrium in Th-series will lead to equal aktivity of <sup>232</sup>Th and <sup>228</sup>Th. Since <sup>232</sup>Th has a incredibly long half-life and <sup>228</sup>Th is short compared to this and we can approximate N(Th)<math>\approx</math>N(<sup>232</sup>Th)=1.25<math>\cdot</math>10^24 The disintegration for both is 1.96<math>\cdot</math>10<sup>6</sup>Bq.  
 
#1000g Th(NO3)4 = 2.083 mol arrow N(Th)= 1.25<math>\cdot</math>10<sup>24</sup> atoms. This is natural thorium, where the equilibrium in Th-series will lead to equal aktivity of <sup>232</sup>Th and <sup>228</sup>Th. Since <sup>232</sup>Th has a incredibly long half-life and <sup>228</sup>Th is short compared to this and we can approximate N(Th)<math>\approx</math>N(<sup>232</sup>Th)=1.25<math>\cdot</math>10^24 The disintegration for both is 1.96<math>\cdot</math>10<sup>6</sup>Bq.  
 
#6.43<math>\cdot</math>10<sup>-8</sup>g  
 
#6.43<math>\cdot</math>10<sup>-8</sup>g  
#10000 Bq <sup>228</sup>Ra = 2.62<math>\cdot</math>10<sup>12</sup> atoms = 90% arrow 100% 2.92<math>\cdot</math>10<sup>12</sup> atoms. If <sup>232</sup>Th is N1 and <sup>228</sup>Ra is N2 we can use the formulas for mother/daughter realations:<math>N_{2}=\frac{\lambda_{1}}{\lambda_{2}-\lambda_{1}}N_{1,0}(e^{-\lambda 1\cdot t}-e^{-\lambda 2\cdot t})</math>&nbsp;<math>N_{1,0}=N_{2}\frac{\lambda_{2}-\lambda_{1}}{\lambda_{1}}\cdot \frac{1}{e^{-\lambda 1\cdot t}-e^{-\lambda 2\cdot t}}=6.25\cdot10^{22}\, atoms</math><math>\frac{6.26\cdot 10^{22}}{6.022\cdot 10^{23}}=0.104 \, mol\cdot480.06 \, g/mol=50 \, g Th(NO_{3})_{4}</math>Alternatively it can be solved by using D(<sup>228</sup>Ra) = 11 111Bq:&nbsp;<math>D_{2}=D_{1}\left( 1-\frac{1}{2}^{t/t_{(1/2)}}\right)\rightarrow D_{1}= \frac{D_{2}}{\left( 1-\frac{1}{2}^{t/t_{(1/2)}}\right)}</math><math>=\frac{11 111\, Bq }{1-\frac{1}{2}^{1/5.75 \, y}}=97838\, Bq</math>
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#10000 Bq <sup>228</sup>Ra = 2.62<math>\cdot</math>10<sup>12</sup> atoms = 90% arrow 100% 2.92<math>\cdot</math>10<sup>12</sup> atoms. If <sup>232</sup>Th is N1 and <sup>228</sup>Ra is N2 we can use the formulas for mother/daughter realations:<math>N_{2}=\frac{\lambda_{1}}{\lambda_{2}-\lambda_{1}}N_{1,0}(e^{-\lambda 1\cdot t}-e^{-\lambda 2\cdot t})</math>&nbsp;<math>N_{1,0}=N_{2}\frac{\lambda_{2}-\lambda_{1}}{\lambda_{1}}\cdot \frac{1}{e^{-\lambda 1\cdot t}-e^{-\lambda 2\cdot t}}=6.25\cdot10^{22}\, atoms</math><math>\frac{6.26\cdot 10^{22}}{6.022\cdot 10^{23}}=0.104 \, mol\cdot480.06 \, g/mol=50 \, g Th(NO_{3})_{4}</math>Alternatively it can be solved by using D(<sup>228</sup>Ra) = 11 111Bq:&nbsp;<math>D_{2}=D_{1}\left( 1-\frac{1}{2}^{t/t_{(1/2)}}\right)\rightarrow D_{1}= \frac{D_{2}}{\left( 1-\frac{1}{2}^{t/t_{(1/2)}}\right)}</math><math>=\frac{11 111\, Bq }{1-\frac{1}{2}^{1/5.75 \, y}}=97838\, Bq</math><math>N_{2}=\frac{D_{2}}{\lambda}=\frac{97838 \, Bq}{\frac{ln 2}{t_{1/2}}}</math>
  
 
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Revision as of 09:19, 19 June 2012

1:

  1. 1000g Th(NO3)4 = 2.083 mol arrow N(Th)= 1.25[math]\cdot[/math]1024 atoms. This is natural thorium, where the equilibrium in Th-series will lead to equal aktivity of 232Th and 228Th. Since 232Th has a incredibly long half-life and 228Th is short compared to this and we can approximate N(Th)[math]\approx[/math]N(232Th)=1.25[math]\cdot[/math]10^24 The disintegration for both is 1.96[math]\cdot[/math]106Bq.
  2. 6.43[math]\cdot[/math]10-8g
  3. 10000 Bq 228Ra = 2.62[math]\cdot[/math]1012 atoms = 90% arrow 100% 2.92[math]\cdot[/math]1012 atoms. If 232Th is N1 and 228Ra is N2 we can use the formulas for mother/daughter realations:[math]N_{2}=\frac{\lambda_{1}}{\lambda_{2}-\lambda_{1}}N_{1,0}(e^{-\lambda 1\cdot t}-e^{-\lambda 2\cdot t})[/math] [math]N_{1,0}=N_{2}\frac{\lambda_{2}-\lambda_{1}}{\lambda_{1}}\cdot \frac{1}{e^{-\lambda 1\cdot t}-e^{-\lambda 2\cdot t}}=6.25\cdot10^{22}\, atoms[/math][math]\frac{6.26\cdot 10^{22}}{6.022\cdot 10^{23}}=0.104 \, mol\cdot480.06 \, g/mol=50 \, g Th(NO_{3})_{4}[/math]Alternatively it can be solved by using D(228Ra) = 11 111Bq: [math]D_{2}=D_{1}\left( 1-\frac{1}{2}^{t/t_{(1/2)}}\right)\rightarrow D_{1}= \frac{D_{2}}{\left( 1-\frac{1}{2}^{t/t_{(1/2)}}\right)}[/math][math]=\frac{11 111\, Bq }{1-\frac{1}{2}^{1/5.75 \, y}}=97838\, Bq[/math][math]N_{2}=\frac{D_{2}}{\lambda}=\frac{97838 \, Bq}{\frac{ln 2}{t_{1/2}}}[/math]