Difference between revisions of "Solutions 2"

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#<sup>228</sup>Ac, <sup>220</sup>Rn,<sup>216</sup>Po, <sup>212</sup>Pb, <sup>212</sup>Bi, <sup>212</sup>Po. <br>
 
#<sup>228</sup>Ac, <sup>220</sup>Rn,<sup>216</sup>Po, <sup>212</sup>Pb, <sup>212</sup>Bi, <sup>212</sup>Po. <br>
  
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'''2:'''<br>
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'''2:'''<br>  
  
#When T= 0 it's only the natural isotopes of uranium: <sup>238</sup>U, <sup>235</sup>U and <sup>234</sup>U.
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#When T= 0 it's only the natural isotopes of uranium: <sup>238</sup>U, <sup>235</sup>U and <sup>234</sup>U.  
#D(<sup>238</sup>U)=D(<sup>234</sup>U)<math>\approx</math> 12.5 kBq, D(<sup>235</sup>U) = 575 Bq.
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#D(<sup>238</sup>U)=D(<sup>234</sup>U)<math>\approx</math> 12.5 kBq, D(<sup>235</sup>U) = 575 Bq.  
#When t = 23.5 h there is created some <sup>234</sup>Th and some <sup>234</sup>Pa, but creation of other daugthers from <sup>238</sup>U is negligible. From the <sup>235</sup>U there is created<sup>231</sup>Th<br>
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#When t = 23.5 h there is created some <sup>234</sup>Th and some <sup>234</sup>Pa, but creation of other daugthers from <sup>238</sup>U is negligible. From the <sup>235</sup>U there is created<sup>231</sup>Th<br>  
 
#D(<sup>238</sup>U) = D(<sup>234</sup>U) =D0, <br>D(<sup>234</sup>Th) = D(<sup>234</sup>Pa) = 376 Bq,<br>D<sup>(235</sup>U) = D0,<br>D(<sup>231</sup>Th) = 287.5 Bq.<br>  
 
#D(<sup>238</sup>U) = D(<sup>234</sup>U) =D0, <br>D(<sup>234</sup>Th) = D(<sup>234</sup>Pa) = 376 Bq,<br>D<sup>(235</sup>U) = D0,<br>D(<sup>231</sup>Th) = 287.5 Bq.<br>  
 
#When t = 23 days the same radionuclides are present.<br>  
 
#When t = 23 days the same radionuclides are present.<br>  
#D(<sup>238</sup>U) = D<sup>(234</sup>U) =D<sub>0</sub>, <br>D(<sup>234</sup>Th) = D(<sup>234</sup>Pa) = 6250 Bq,<br>D(<sup>235</sup>U) = D<sub>0</sub>,<br>D(<sup>231</sup>Th) = D(<sup>235</sup>U) =575 Bq.&nbsp;<br>
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#D(<sup>238</sup>U) = D<sup>(234</sup>U) =D<sub>0</sub>, <br>D(<sup>234</sup>Th) = D(<sup>234</sup>Pa) = 6250 Bq,<br>D(<sup>235</sup>U) = D<sub>0</sub>,<br>D(<sup>231</sup>Th) = D(<sup>235</sup>U) =575 Bq.&nbsp;<br>  
#&nbsp;when t = 1.0 y the same radionuclides are present.<br>
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#When t = 1.0 y the same radionuclides are present.<br>  
##&nbsp;<br>
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#D(<sup>238</sup>U) = D(<sup>234</sup>U) =D<sub>0</sub>, <br>D(<sup>234</sup>Th) = D(<sup>234</sup>Pa) = D<sup>(238</sup>U) = 12,5 kBq,<br>D(<sup>235</sup>U) = D<sub>0</sub>,<br>D(<sup>231</sup>Th) = D(<sup>235</sup>U) =575 Bq.<br>
#&nbsp;<br>
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#When t = 1.0 y the same radionuclides are present.<br>
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#&nbsp;<br>  
 
#<br>
 
#<br>
  
 
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Revision as of 09:54, 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}}}=6.26 \cdot 10^{22} \, atoms[/math]
  4. 224Ra is created from 228Th immeasurable amounts of 228Th is created in three days, creation of new 224Ra can therefore be ignored. D0(224Ra)=D0(228Th)=1.36[math]\cdot[/math]106 Bg, and we get a normal decay:                                      [math]D=D_{0}\cdot e^{-\lambda t}=1.1\cdot 10^{6} \,Bq[/math]
  5. 228Ac, 220Rn,216Po, 212Pb, 212Bi, 212Po.


2:

  1. When T= 0 it's only the natural isotopes of uranium: 238U, 235U and 234U.
  2. D(238U)=D(234U)[math]\approx[/math] 12.5 kBq, D(235U) = 575 Bq.
  3. When t = 23.5 h there is created some 234Th and some 234Pa, but creation of other daugthers from 238U is negligible. From the 235U there is created231Th
  4. D(238U) = D(234U) =D0,
    D(234Th) = D(234Pa) = 376 Bq,
    D(235U) = D0,
    D(231Th) = 287.5 Bq.
  5. When t = 23 days the same radionuclides are present.
  6. D(238U) = D(234U) =D0,
    D(234Th) = D(234Pa) = 6250 Bq,
    D(235U) = D0,
    D(231Th) = D(235U) =575 Bq. 
  7. When t = 1.0 y the same radionuclides are present.
  8. D(238U) = D(234U) =D0,
    D(234Th) = D(234Pa) = D(238U) = 12,5 kBq,
    D(235U) = D0,
    D(231Th) = D(235U) =575 Bq.
  9. When t = 1.0 y the same radionuclides are present.
  10.