Difference between revisions of "Solutions 2"
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====== Written and developed by [http://www.mn.uio.no/kjemi/english/people/aca/phoff/index.html Prof. Per Hoff] (uio) ====== | ====== Written and developed by [http://www.mn.uio.no/kjemi/english/people/aca/phoff/index.html Prof. Per Hoff] (uio) ====== | ||
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Revision as of 09:58, 22 June 2012
Written and developed by Prof. Per Hoff (uio)
Return to Problem Solving Sets
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