Question

The half life of radium -226 is 1640 years. If a sample contains 200 mg, how many mg will remain after 4000 years?

Answer 1

`\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 \\ (N)/(N_o)=e^(\lambda t) \\ \\ \ln ((N)/(N_o))=\ln (e^(\lambda t)) \\ \\ \ln ((N)/(N_o))=\lambda t \\ (N)/(N_o)=(1)/(2)=0.5 \\ \ln (0.5)=\lambda t \\ \lambda=(\ln(0.5))/(t) \\ \\ \lambda=(\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}`