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(A) The radioactive decay law is given by $N(t) = N_{0} e^{-\lambda t}$,where $\lambda = \frac{\ln 2}{T}$.Let $t_{1}$ be the time when the number of a

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$T \frac{\log 5}{\log 2}$

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(A) The radioactive decay law is given by $N(t) = N_{0} e^{-\lambda t}$,where $\lambda = \frac{\ln 2}{T}$.Let $t_{1}$ be the time when the number of atoms is $N_{0}/2$ and $t_{2}$ be the time when the number of atoms is $N_{0}/10$.For $N(t_{1}) = N_{0}/2$,we have $\frac{N_{0}}{2} = N_{0} e^{-\lambda t_{1}} \Rightarrow e^{\lambda t_{1}} = 2 \Rightarrow \lambda t_{1} = \ln 2$.For $N(t_{2}) = N_{0}/10$,we have $\frac{N_{0}}{10} = N_{0} e^{-\lambda t_{2}} \Rightarrow e^{\lambda t_{2}} = 10 \Rightarrow \lambda t_{2} = \ln 10$.The time interval required is $\Delta t = t_{2} - t_{1} = \frac{\ln 10 - \ln 2}{\lambda} = \frac{\ln(10/2)}{\lambda} = \frac{\ln 5}{\lambda}$.Since $\lambda = \frac{\ln 2}{T}$,we substitute $\lambda$ to get $\Delta t = \frac{\ln 5}{(\ln 2 / T)} = T \frac{\ln 5}{\ln 2} = T \log_{2} 5$.

The number of atoms of a radioactive substance of half-life $T$ is $N_{0}$ at $t=0$. The time necessary to decay from $N_{0} / 2$ atoms to $N_{0} / 10$ atoms will be

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