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  • The average life of a radioactive element is 17.2 min Calculate the time interval (in seconds) between the stages of 33.33\% and 50\% decay? (almost)Option: 1</s

The average life of a radioactive element is 17.2 min Calculate the time interval (in seconds) between the stages of 33.33\% and 50\% decay? (almost)

Option: 1

200 s


Option: 2

50 s


Option: 3

267 s


Option: 4

297 s


Answers (1)

best_answer

\mathrm{\begin{aligned} & \lambda=\frac{1}{\tau}=\frac{1}{t} \ln \frac{\left[R_0\right]}{\left[R_t\right]} \\ & \frac{1}{17.2}=\frac{1}{t_1} \ln \frac{100}{100-\frac{100}{2}} \text {. } \\ & t_1=17.2 \ln 2 \\ & \frac{1}{17.2}=\frac{1}{t_2} \ln \frac{100}{100-\frac{100}{3}} \\ & t_2=17.2 \ln \frac{3}{2}=17.2 \ln 3-17.2 \ln 2 \\ & t_1-t_2=2 \times 17.2 \ln 2-17.2 \ln 3 \\ & =4.95 \mathrm{~min} \\ & =4.95 \times 60=296.8 \mathrm{sgc} . \\ & \end{aligned}}\mathrm{\begin{aligned} & \lambda=\frac{1}{\tau}=\frac{1}{t} \ln \frac{\left[R_0\right]}{\left[R_t\right]} \\ & \frac{1}{17.2}=\frac{1}{t_1} \ln \frac{100}{100-\frac{100}{2}} \text {. } \\ & t_1=17.2 \ln 2 \\ & \frac{1}{17.2}=\frac{1}{t_2} \ln \frac{100}{100-\frac{100}{3}} \\ & t_2=17.2 \ln \frac{3}{2}=17.2 \ln 3-17.2 \ln 2 \\ & t_1-t_2=2 \times 17.2 \ln 2-17.2 \ln 3 \\ & =4.95 \mathrm{~min} \\ & =4.95 \times 60=296.8 \ \mathrm{sec} . \\ & \end{aligned}}

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