Solutions 2

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  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]