For a radioactive element,the mean life is ta | vedclass

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(C) The activity $I$ at any time $t$ is given by $I(t) = I_0 e^{-\lambda t}$,where $I_0$ is the initial activity at $t=0$.Given $I_0 = n$ and the mean

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$n	au \left( {1 - {e^{ - \frac{t}{	au }}}} ight)$

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(C) The activity $I$ at any time $t$ is given by $I(t) = I_0 e^{-\lambda t}$,where $I_0$ is the initial activity at $t=0$.Given $I_0 = n$ and the mean life $	au = 1/\lambda$,we have $\lambda = 1/	au$.The number of nuclei decaying in the time interval $0$ to $t$ is the integral of the activity over that time:$N_{decayed} = \int_{0}^{t} I(t') dt' = \int_{0}^{t} I_0 e^{-\lambda t'} dt'$$N_{decayed} = I_0 \left[ \frac{e^{-\lambda t'}}{-\lambda} ight]_{0}^{t} = \frac{I_0}{\lambda} (1 - e^{-\lambda t})$Substituting $I_0 = n$ and $\lambda = 1/	au$:$N_{decayed} = n 	au (1 - e^{-t/	au})$.

For a radioactive element,the mean life is $\tau$. At time $t = 0$,the number of nuclei decaying per unit time is $n$. The number of nuclei that have decayed between time $0$ and $t$ is:

Similar Questions

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If the half-life of radium is $77$ days,its decay constant in $day^{-1}$ will be:

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