The activity of a radioactive sample is I_0 c | vedclass

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(A) Given that at $t = 0$,activity $I = I_0$ and at $t = 5$ minutes,$I = I_0/e$.Using the radioactive decay law $I = I_0 e^{-\lambda t}$:$\frac{I_0}{e

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$5 \ln(2)$

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(A) Given that at $t = 0$,activity $I = I_0$ and at $t = 5$ minutes,$I = I_0/e$.Using the radioactive decay law $I = I_0 e^{-\lambda t}$:$\frac{I_0}{e} = I_0 e^{-5\lambda}$$e^{-1} = e^{-5\lambda}$Comparing the exponents,$5\lambda = 1$,so $\lambda = \frac{1}{5} 	ext{ min}^{-1}$.The time $t$ at which the activity becomes half of its initial value is the half-life $T_{1/2}$.$T_{1/2} = \frac{\ln(2)}{\lambda} = \frac{\ln(2)}{1/5} = 5 \ln(2)$.Since $\ln(2) = \log_e(2)$,the answer is $5 \log_e(2)$.

The activity of a radioactive sample is $I_0$ counts/minute at $t = 0$ and $I_0/e$ counts/minute at $t = 5$ minutes. At what time (in minutes) will its activity decrease to half of its initial value?

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