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(C) The number of active nuclei present in the original sample is $N_0 = 4 	imes 10^{16}$.The half-life of the element is $T_{1/2} = 10$ days.The num

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$3.5$

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(C) The number of active nuclei present in the original sample is $N_0 = 4 	imes 10^{16}$.The half-life of the element is $T_{1/2} = 10$ days.The number of half-lives in $30$ days is $n = \frac{30}{10} = 3$.The number of active nuclei remaining after $30$ days is $N = \frac{N_0}{2^n} = \frac{4 	imes 10^{16}}{2^3} = \frac{4 	imes 10^{16}}{8} = 0.5 	imes 10^{16}$.The number of decayed nuclei is given by $N_{decayed} = N_0 - N$.$N_{decayed} = 4 	imes 10^{16} - 0.5 	imes 10^{16} = 3.5 	imes 10^{16}$.

$A$ sample of a radioactive element contains $4 \times 10^{16}$ active nuclei. If the half-life of the element is $10$ days,then the number of decayed nuclei after $30$ days is ........ $\times 10^{16}$.

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