1.
[tex]Dane:\\T = 2,2\cdot10^{6} \ lat\\N_{o} = 200 \ g\\t = 6,6\cdot10^{8} \ lat\\Szukane:\\N(t) = ?[/tex]
Z równania rozpadu promieniotwórczego:
[tex]N(t) = N_{o}\cdot(\frac{1}{2})^{\frac{t}{T}}\\\\N(t) = 200 \ g\cdot(\frac{1}{2})^{\frac{6,6\cdot10^{6}}{2,2\cdot10^{6}}}\\\\N(t) = 200 \ g\cdot(\frac{1}{2})^{3}\\\\N(t) = 200 \ g\cdot\frac{1}{8}\\\\\boxed{N(t) = 25 \ g}[/tex]
Odp. Pozostanie 25 g pierwiastka.
2.
[tex]T = 2\cdot10^{8} \ lat\\N_{o} = 1 \ kg\\t_1 = 4\cdot10^{8} \ lat\\t_2 = 10^{8} \ lat\\Szukane:\\N(t_1), N(t_2) = ?\\\\\\N(t) = N_{o}\cdot(\frac{1}{2})^{\frac{t}{T}}\\\\N(t_1) = N_{o}\cdot(\frac{1}{2})^{\frac{t_1}{T}}\\\\N(t_1) = 1 \ kg\cdot(\frac{1}{2})^{\frac{4\cdot10^{8}}{2\cdot10^{8}}}\\\\N(t_1)=1 \ kg\cdot(\frac{1}{2})^{2}\\\\N(t_1) = 1 \ kg\cdot\frac{1}{4}\\\\\boxed{N(t_1) = 0,25 \ kg}[/tex]
[tex]N(t_2) = N_{o}\cdot(\frac{1}{2})^{\frac{t_2}{T}}}\\\\N(t_2) = 1 \ kg\cdot(\frac{1}{2})^{\frac{10^{8}}{2\cdot10^{8}}}\\\\N(t_2) = 1 \ kg\cdot(\frac{1}{2})^{\frac{1}{2}}\\\\N(t_2) = 1 \ kg\cdot\sqrt{\frac{1}{2}}\\\\N(t_2) = 1 \ kg \cdot0,707\\\\\boxed{N(t_2) \approx 0,707 \ kg}[/tex]
Odp. Po czasie t₁ pozostanie 0,25 kg pierwiastka, a po czasie t₂ - 0,707 kg.
