If N_t = N_o e^{-lambda t},then the number of | vedclass

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(C) The number of undecayed nuclei at any time $t$ is given by $N_t = N_o e^{-\lambda t}$.At time $t_1$,the number of undecayed nuclei is $N(t_1) = N_

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$N_o [e^{-\lambda t_1} - e^{-\lambda t_2}]$

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(C) The number of undecayed nuclei at any time $t$ is given by $N_t = N_o e^{-\lambda t}$.At time $t_1$,the number of undecayed nuclei is $N(t_1) = N_o e^{-\lambda t_1}$.At time $t_2$,the number of undecayed nuclei is $N(t_2) = N_o e^{-\lambda t_2}$.The number of nuclei that have decayed between time $t_1$ and $t_2$ is the difference between the number of undecayed nuclei at $t_1$ and $t_2$:$\Delta N = N(t_1) - N(t_2)$Substituting the expressions:$\Delta N = N_o e^{-\lambda t_1} - N_o e^{-\lambda t_2}$Taking $N_o$ as a common factor:$\Delta N = N_o [e^{-\lambda t_1} - e^{-\lambda t_2}]$

If $N_t = N_o e^{-\lambda t}$,then the number of nuclei decaying between time $t_1$ and $t_2$ $(t_2 > t_1)$ is given by:

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