Solutions 2

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 = 10.0 y the same radionuclides are present.
   
10.  same as 8.
    

3:

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 this gives: [math]D_{(^{226}Ra)}=D_{(^{238}U)}=125\,kBq \rightarrow N=\frac{D}{\lambda}=9.1 \cdot 10^{15} = 3.4\cdot10^{-5}\, 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^{-5}\, 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 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:

1. It is easily accessible, it needs to be processed once to create many doses of medicine.
2. The following nuclides can be extracted from a nuclide extracted: ⁶⁸Ga, ⁹⁰Y, ²¹²Pb.
3. 100 Mbq ²⁰¹Tl = 9.11[math]\cdot[/math]10¹⁴ atoms = 3[math]\cdot[/math]10^-7g.
4. The amount inserted is so small it is not possible for it to be poisonous for humans.
   

5:

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