Difference between revisions of "Solutions 2"
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− | #1000g Th( | + | #1000g Th(NO<sub>3</sub>)<sub>4</sub> = 2.083 mol<math>\rightarrow</math>N(Th)= 1.25 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(232Th)=1.25<math>\cdot</math>10<sup>24</sup> The disintegration for both is 1.96<math>\cdot</math>10<sup>6</sup>Bq.<br> |
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#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> <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: <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> | #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> <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: <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> | ||
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#D<sub>1</sub>(<sup>44</sup>Ti) ≈ D<sub>0</sub>(<sup>44</sup>Ti) = 100 MBq, m(<sup>44</sup>Ti) = 2<math>\cdot</math>10<sup>-5</sup> g<br>D<sub>1</sub>(<sup>44</sup>Sc) = 50 MBq, m(<sup>44</sup>Sc) = 7.4<math>\cdot</math>10<sup>-11</sup> g<br> | #D<sub>1</sub>(<sup>44</sup>Ti) ≈ D<sub>0</sub>(<sup>44</sup>Ti) = 100 MBq, m(<sup>44</sup>Ti) = 2<math>\cdot</math>10<sup>-5</sup> g<br>D<sub>1</sub>(<sup>44</sup>Sc) = 50 MBq, m(<sup>44</sup>Sc) = 7.4<math>\cdot</math>10<sup>-11</sup> g<br> | ||
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− | #The gamma radiation comes from the daughters <sup>228</sup>Ac and <sup>208</sup>Tl. | + | #The gamma radiation comes from the daughters <sup>228</sup>Ac and <sup>208</sup>Tl. |
− | #Thorium decay series we find <sup>232</sup>Th and <sup>228</sup>Th. They exist in nature in equalibrum so that D<sub>1</sub>=D<sub>2</sub> and there is 7.34<math>\cdot</math>10<sup>9</sup> more mass of <sup>232</sup>Th than <sup>228</sup>Th. | + | #Thorium decay series we find <sup>232</sup>Th and <sup>228</sup>Th. They exist in nature in equalibrum so that D<sub>1</sub>=D<sub>2</sub> and there is 7.34<math>\cdot</math>10<sup>9</sup> more mass of <sup>232</sup>Th than <sup>228</sup>Th. |
#To achive equalibrium trough the whole series it needs to have taken ten times longer than the most longlived daughter; for <sup>238</sup>U this is 2.455<math>\cdot</math>10<sup>6</sup> y for <sup>235</sup>U it is 3.276<math>\cdot</math>10<sup>5</sup> y and for <sup>232</sup>Th 57.5 y | #To achive equalibrium trough the whole series it needs to have taken ten times longer than the most longlived daughter; for <sup>238</sup>U this is 2.455<math>\cdot</math>10<sup>6</sup> y for <sup>235</sup>U it is 3.276<math>\cdot</math>10<sup>5</sup> y and for <sup>232</sup>Th 57.5 y | ||
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− | #100 g natural Th is more or less only <sup>232</sup>Th, this gives a decay of 405.9 kBq. D(<sup>229</sup>Th) = D(232Th), D(total) = 811.8 kBq. | + | #100 g natural Th is more or less only <sup>232</sup>Th, this gives a decay of 405.9 kBq. D(<sup>229</sup>Th) = D(232Th), D(total) = 811.8 kBq. |
− | #405.9 kBq is 1.34<math>\cdot</math> 10<sup>-8</sup> g. | + | #405.9 kBq is 1.34<math>\cdot</math> 10<sup>-8</sup> g. |
− | #nothing – <sup>229</sup>Th does not exist in nature. | + | #nothing – <sup>229</sup>Th does not exist in nature. |
#After 7.2 days the activity of <sup>228</sup>Ra will be 963 Bq and it will be in equilibrium with <sup>228</sup>Ac. From <sup>228</sup>Th it is created <sup>224</sup>Ra after two half-lifes there will be 75% of max possible 224Ra and in equalibrium with this<sup>210</sup>Rn, <sup>216</sup>Po, <sup>212</sup>Pb, <sup>212</sup>Bi (assume 50% branching to <sup>208</sup>Tl and <sup>212</sup>Po). Total alpha activity: 2.2029 MBq, beta-activety: 610.7 kBq. | #After 7.2 days the activity of <sup>228</sup>Ra will be 963 Bq and it will be in equilibrium with <sup>228</sup>Ac. From <sup>228</sup>Th it is created <sup>224</sup>Ra after two half-lifes there will be 75% of max possible 224Ra and in equalibrium with this<sup>210</sup>Rn, <sup>216</sup>Po, <sup>212</sup>Pb, <sup>212</sup>Bi (assume 50% branching to <sup>208</sup>Tl and <sup>212</sup>Po). Total alpha activity: 2.2029 MBq, beta-activety: 610.7 kBq. | ||
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− | + | #5.78 hours. | |
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− | #5.78 hours. | ||
#After two half-lives, 76 hours, there will be 75%, 7500 Bq. | #After two half-lives, 76 hours, there will be 75%, 7500 Bq. | ||
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− | + | #The alpha decay of <sup>211</sup>At gives <sup>207</sup>Bi, with a half-life of 31.55 years. | |
− | + | #6.57<math>\cdot</math>10<sup>-7</sup> | |
− | #The alpha decay of <sup>211</sup>At gives <sup>207</sup>Bi, with a half-life of 31.55 years. | ||
− | #6.57<math>\cdot</math>10<sup>-7</sup> | ||
#after a week all of the <sup>211</sup>At will decay to <sup>207</sup>Bi. The half-life of this is so long (31.55 years) that we can do the approximation N(<sup>207</sup>Bi)<math>\approx</math>N<sub>0</sub>(<sup>211</sup>At)=1.87<math>\cdot</math>10<sup>15</sup> | #after a week all of the <sup>211</sup>At will decay to <sup>207</sup>Bi. The half-life of this is so long (31.55 years) that we can do the approximation N(<sup>207</sup>Bi)<math>\approx</math>N<sub>0</sub>(<sup>211</sup>At)=1.87<math>\cdot</math>10<sup>15</sup> | ||
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Revision as of 11:25, 19 June 2012
1:
- 1000g Th(NO3)4 = 2.083 mol
- 6.43 10-8g
- 10000 Bq 228Ra = 2.62 1012 atoms = 90% arrow 100% 2.92 1012 atoms. If 232Th is N1 and 228Ra is N2 we can use the formulas for mother/daughter realations: Alternatively it can be solved by using D(228Ra) = 11 111Bq:
- 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
106 Bg, and we get a normal decay: - 228Ac, 220Rn,216Po, 212Pb, 212Bi, 212Po.
2:
- When T= 0 it's only the natural isotopes of uranium: 238U, 235U and 234U.
- D(238U)=D(234U) 12.5 kBq, D(235U) = 575 Bq.
- 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
- D(238U) = D(234U) =D0,
D(234Th) = D(234Pa) = 376 Bq,
D(235U) = D0,
D(231Th) = 287.5 Bq.
- When t = 23 days the same radionuclides are present.
- D(238U) = D(234U) =D0,
D(234Th) = D(234Pa) = 6250 Bq,
D(235U) = D0,
D(231Th) = D(235U) =575 Bq.
- When t = 1.0 y the same radionuclides are present.
- D(238U) = D(234U) =D0,
D(234Th) = D(234Pa) = D(238U) = 12,5 kBq,
D(235U) = D0,
D(231Th) = D(235U) =575 Bq.
- When t = 10.0 y the same radionuclides are present.
- same as 8.
3:
- The shale contains all of the daughter products from 238U and 235U in equilibrium. In 10 g natural Uranium there is 125 kBq 238U and 5.75 kBq 235U this gives:
- 210Pb exist as a daughter from 238U:
- One of the daughters is 210Po, which is a alpha emitter and it is can do great harm if it gets inside the body. In addition Pb is a daughter of radon which makes it possible for it to enter the lungs.
4:
- It is easily accessible, it needs to be processed once to create many doses of medicine.
- The following nuclides can be extracted from a nuclide extracted: 68Ga, 90Y, 212Pb.
- 100 Mbq 201Tl = 9.11 1014 atoms = 3 10-7g.
- The amount inserted is so small it is not possible for it to be poisonous for humans.
5:
- After one half-life it will be 50 Mbq so 3.92 h.
- D1(44Ti) ≈ D0(44Ti) = 100 MBq, m(44Ti) = 2
D1(44Sc) = 50 MBq, m(44Sc) = 7.4 10-11 g 10-5 g
6:
- The gamma radiation comes from the daughters 228Ac and 208Tl.
- Thorium decay series we find 232Th and 228Th. They exist in nature in equalibrum so that D1=D2 and there is 7.34 109 more mass of 232Th than 228Th.
- To achive equalibrium trough the whole series it needs to have taken ten times longer than the most longlived daughter; for 238U this is 2.455 106 y for 235U it is 3.276 105 y and for 232Th 57.5 y
7:
- 100 g natural Th is more or less only 232Th, this gives a decay of 405.9 kBq. D(229Th) = D(232Th), D(total) = 811.8 kBq.
- 405.9 kBq is 1.34 10-8 g.
- nothing – 229Th does not exist in nature.
- After 7.2 days the activity of 228Ra will be 963 Bq and it will be in equilibrium with 228Ac. From 228Th it is created 224Ra after two half-lifes there will be 75% of max possible 224Ra and in equalibrium with this210Rn, 216Po, 212Pb, 212Bi (assume 50% branching to 208Tl and 212Po). Total alpha activity: 2.2029 MBq, beta-activety: 610.7 kBq.
8:
- 5.78 hours.
- After two half-lives, 76 hours, there will be 75%, 7500 Bq.
9:
- The alpha decay of 211At gives 207Bi, with a half-life of 31.55 years.
- 6.57 10-7
- after a week all of the 211At will decay to 207Bi. The half-life of this is so long (31.55 years) that we can do the approximation N(207Bi) N0(211At)=1.87 1015