Respuesta :
[tex]\begin{gathered} The\text{ following function } \\ N=N_oe^{\lambda t}\text{, N=}amount\text{ of radium-226} \\ t=\text{ time} \\ \lambda\text{= constant} \\ \text{Radium -226} \\ \text{half life = 1640 years} \\ U\sin g\text{ last information, we can find the value of }\lambda \\ \frac{N}{N_o}=e^{\lambda t} \\ \\ \ln (\frac{N}{N_o})=\ln (e^{\lambda t}) \\ \\ \ln (\frac{N}{N_o})=\lambda t \\ \frac{N}{N_o}=\frac{1}{2}=0.5 \\ \ln (0.5)=\lambda t \\ \lambda=\frac{\ln(0.5)}{t} \\ \\ \lambda=\frac{\ln(0.5)}{1640} \\ \\ \lambda=-0.000423 \\ Now,\text{ on the case of 200mg of radium -226} \\ N=200e^{-0.000423t} \\ t=4000\text{ years} \\ \\ N=200e^{-0.000423\cdot(4000)} \\ N=36.83 \\ \text{After 4000 years the sample will contain }36.83mg\text{ of radium -226} \end{gathered}[/tex]
