Suggested+Solutions+to+Mother+Daughter+Relations+and+Equilibrium

1: > 2.083 mole ⟶ N(Th) > 1.25•10 24 atoms. This is natural thorium, where the equilibrium in Th-series will lead to equal activity of 232 Th and 228 Th. Since 232 Th has a incredibly long half-life and 228 Th is short compared to this and we can approximate N(Th)≈N 232 Th)=1.25•10 24 The disintegration for both is 1.96•10 6 Bq. math N_2=\frac{\lambda_1}{\lambda_2-\lambda_1}N_{1,0}\left(e^{\lambda_1 t}-e^{-\lambda_2 t} \right) math math N_{1,0}=N_2\frac{\lambda_2-\lambda_1}{\lambda_1}\cdot\frac{1}{e^{-\lambda_1 t}-e^{\lambda_2 t}}=6.25\cdot10^{22} atoms math math \frac{6.26\cdot10^{22}}{6.022\cdot10^23}=0.104mol\cdot480.06g/mol=50\,g\,Th(NO_3)_4 math
 * 1) 1000g Th(NO 3 ) 4
 * 1) 6.46•10 -8 g
 * 2) 10000 Bq 228 Ra = 2.62•10 12 atoms = 90%⟶100% 2.91•10 12 atoms. If 232 >Th is N 1 and 228 Ra is N 2 we can use the formulas for mother/daughter relations:

lternatively it can be solved by using D( 228 Ra) = 11 111Bq: math D_2=D_1\left(1-\frac{1}{2}^{t/t_{(1/2)}}\right)\rightarrow D_1=\frac{D_2}{1-\frac{1}{2}^{t/t_{(1/2)}}} math math =\frac{11 111\, Bq }{1-\frac{1}{2}^{1/5.75 \, y}}=97838\, Bq math

math D=D_{0}\cdot e^{-\lambda t}=1.1\cdot 10^{6} \,Bq math
 * 1) 228 Ra is formed from 228 Th, an immeasurable amount of 228 Th is formed in three days, and formation of new 224 Ra can therefore be ignored. D 0 ( 224 Ra)=D 0 ( 228 Th)=1.36•10 6 Bq. and we get a normal decay:
 * 1) 228 Ac, 220 Rn, 216 Po, 212 Pb, 212 Bi, 212 Po.


 * 2:**

> D( 235 U) = D0, > D( 231 Th) = 287.5 Bq. > D( 234 Th) = D( 234 Pa) = 6250 Bq, > D( 235 U) = D0, > D( 231 Th) = D( 235 U) =575 Bq. > D( 234 Th) = D( 234 Pa) = D( 238 U) = 12,5 kBq, > D( 235 U) = D 0 , > D( 231 Th) = D( 235 U) =575 Bq.
 * 1) When T= 0 only the natural isotopes of uranium is present: 238 U, 235 U and 234 U.
 * 2) D/(span style="font-size: 75%; vertical-align: super;">238U)=D( 234 U) ≈ 12.5 kBq, D( 235 U) = 575 Bq.
 * 3) When t = 23.5 h there is created some 234 Th and some 234 Pa, but creation of other daughters from 238 U is negligible. From the 235 U there is created 231 Th
 * 4) D( 238 U) = D( 234 U) =D 0 ,> D( 234 Th) = D( 234 Pa) = 376 Bq,
 * 1) When t = 23 days the same radionuclides are present.
 * 2) D( 238 U) = D( 234 U) =D 0 ,
 * 1) When t = 1.0 y the same radionuclides are present.
 * 2) D( 238 U) = D( 234 U) =D 0 ,
 * 1) When t = 10.0 y the same radionuclides are present.
 * 2) same as 8.


 * 3:**

1: The shale contains all of the daughter products from 238 U and 235 U in equilibrium. In 10 g natural Uranium there is 125 kBq 238 U and 5.75 kBq 235 U 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: 210 Pb exist as a daughter from 238 U:

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 210 Po, 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.


 * 4:**
 * 1) It is easily accessible; it only needs to be processed once to create several doses of medicine.
 * 2) The following nuclides can be extracted from a nuclide generator: 68 Ga, 90 Y, 212 Pb.
 * 3) 100 Mbq 201 Tl = 3.79•10 13 atoms → m = 1.27•10 -8 g.
 * 4) The amount inserted is too small to be considered poisonous for humans.

D 1 ( 44 Sc) = 50 MBq, m( 44 Sc) = 7.4•10 -11 g.
 * 5:**
 * 1) After one half-life it will be 50 Mbq, which is 3.92 h ( 44 Sc).
 * 2) D 1 ( 44 Ti) ≈ D 0 ( 44 Ti) = 100 MBq, m( 44 Ti) = 2.01•10 -5 g


 * 6:**
 * 1) The gamma radiation comes from the daughters 228 Ac and 208 Tl.
 * 2) There are 2 radioisotopes in natural Thorium, namely 232/228 Th. 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•10 9 more mass of 232 Th than 228 Th.
 * 3) To achieve equilibrium through the whole series it needs to have taken ten times longer than the most long-lived daughter; for 238 U this is 2.455•10 6 y for 235 U it is 3.276•10 5 y and for 232 Th 57.5 y


 * 7:**
 * 1) 100 g natural Th is more or less only 232 Th, this gives a decay of 405.9 kBq. D( 229 Th) = D( 232 Th), D(total) = 811.8 kBq.
 * 2) 405.9 kBq is 1.34•10 -8 g.
 * 3) Nothing – 229 Th does not exist in nature.
 * 4) After 7.2 days the activity of 228 Ra will be 963 Bq and it will be in equilibrium with 228 Ac. 224 Ra is formed from 228 Th and after two half-lives there will be 75% of max possible 224 Ra in equilibrium with 210 Rn, 216 Po, 212 Pb, 212 Bi (assume 50% branching to 208 Tl and 212 Po). Total alpha activity: 2.2029 MBq, beta-activity: 610.7 kBq.


 * 8:**
 * 1) 5.78 hours.
 * 2) After two half-lives, 76 hours, there will be 75%, 7500 Bq.


 * 9:**
 * 1) The alpha decay of 211 At gives 207 Bi, with a half-life of 31.55 years.
 * 2) 6.57•10 -7 g
 * 3) After one week all of the 211 At will decay to 207 Bi. The half-life is long enough ( 31.55 years) to do the approximation N( 207 Bi) ≈ N 0 ( 211 At)=1.87•10 15


 * 7:**
 * 1) 100 g natural Th is more or less only 232 Th, this gives a decay of 405.9 kBq. D( 229 Th) = D( 232 Th), D(total) = 811.8 kBq.
 * 2) 405.9 kBq is 1.34•10 -8 g.
 * 3) Nothing – 229 Th does not exist in nature.
 * 4) After 7.2 days the activity of 228 Ra will be 963 Bq and it will be in equilibrium with 228 Ac. 224 Ra is formed from 228 Th and after two half-lives there will be 75% of max possible 224 Ra in equilibrium with 210 Rn, 216 Po, 212 Pb, 212 Bi (assume 50% branching to 208 Tl and 212 Po). Total alpha activity: 2.2029 MBq, beta-activity: 610.7 kBq.

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