Solutions 2

From mn/safe/nukwik
Jump to: navigation, search

Mother Daughter Relations and Equilibrium

Return to Problem Solving Sets


  1. 1000g Th(NO3)4 = 2.083 mol[math]\rightarrow[/math]N(Th)= 1.25 1024 atoms. This is natural thorium, where the equilibrium in Th-series will lead to equal activity 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)N(232Th)=1.25[math]\cdot[/math]1024 The disintegration for both is 1.96[math]\cdot[/math]106Bq.
  2. 6.46 [math]\cdot[/math]10-8g
  3. 10000 Bq 228Ra = 2.62 [math]\cdot[/math]1012 atoms = 90% [math]\rightarrow[/math] 100% 2.91 [math]\cdot[/math]1012 atoms. If 232Th is N1 and 228Ra is N2 we can use the formulas for mother/daughter relations:[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. 228Ra is formed from 228Th, Immeasurable amounts of 228Th is formed in three days, formation 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.


  1. When T= 0 only the natural isotopes of uranium is present: 238U, 235U and 234U.
  2. D(238U)=D(234U) ≈ 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 daughters 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 = 10.0 y the same radionuclides are present.
  10.  same as 8.


  1. 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 which gives: [math]D_{(^{226}Ra)}=D_{(^{238}U)}=125\,kBq \rightarrow N=\frac{D}{\lambda}=9.1 \cdot 10^{15} = 3.4\cdot10^{-6}\, g[/math][math]D_{(^{223}Ra)}=D_{(^{235}U)}=5.75\,kBq \rightarrow N=\frac{D}{\lambda}=8.2 \cdot 10^{9} = 3.0\cdot10^{-12}\, g[/math]
  2. 210Pb exist as a daughter from 238U: [math]D_{(^{210}Pb)}=D_{(^{238}U)}=125 kBq \rightarrow N=\frac{D}{\lambda}=1.3\cdot 10^{14}=4.4\cdot10^{-8}\, g[/math]
  3. One of the daughters is 210Po, which is a alpha emitter and can do great harm inside the body. In addition Pb is a daughter of radon which makes it possible for it to enter the lungs.


  1. It is easily accessible, it only needs to be processed once to create many doses of medicine.
  2. The following nuclides can be extracted from a nuclide generator: 68Ga, 90Y, 212Pb.
  3. 100 Mbq 201Tl = 3.79 [math]\cdot[/math]1013 atoms [math]\rightarrow[/math] m = 1.27[math]\cdot[/math]10-8g.
  4. The amount inserted is too small to be considered poisonous for humans.


  1. After one half-life it will be 50 Mbq, which is 3.92 h (44Sc).
  2. D1(44Ti) ≈ D0(44Ti) = 100 MBq, m(44Ti) = 2.01 [math]\cdot[/math]10-5 g
    D1(44Sc) = 50 MBq, m(44Sc) = 7.4 [math]\cdot[/math]10-11 g


  1. The gamma radiation comes from the daughters 228Ac and 208Tl.
  2. There are 2 radioisotopes in natural Thorium, namely 232/228Th. They are in a secular equilibrium and the activity of the latter will be equal to the activity of the former. It is about 7.34 [math]\cdot[/math]109 more mass of 232Th than 228Th. 
  3. 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 [math]\cdot[/math]106 y for 235U it is 3.276[math]\cdot[/math]105 y and for 232Th 57.5 y


  1. 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.
  2. 405.9 kBq is 1.34[math]\cdot[/math] 10-8 g.
  3. nothing – 229Th does not exist in nature.
  4. After 7.2 days the activity of 228Ra will be 963 Bq and it will be in equilibrium with 228Ac. 224Ra is formed from 228Th and after two half-lives there will be 75% of max possible 224Ra in equilibrium with 210Rn, 216Po, 212Pb, 212Bi (assume 50% branching to 208Tl and 212Po). Total alpha activity: 2.2029 MBq, beta-activety: 610.7 kBq.


  1. 5.78 hours.
  2. After two half-lives, 76 hours, there will be 75%, 7500 Bq.


  1. The alpha decay of 211At gives 207Bi, with a half-life of 31.55 years.
  2. 6.57[math]\cdot[/math]10-7g
  3. After one week all of the 211At will decay to 207Bi. The half-life is long enough (31.55 years) to do the approximation N(207Bi) ≈ N0(211At)=1.87[math]\cdot[/math]1015