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(C) The number of nuclei remaining after time $t$ is given by $N = N_0 \left( \frac{1}{2} ight)^n$,where $n = \frac{t}{T_{1/2}}$.Given $N_0 = 4 	im

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$3.5 	imes 10^{16}$

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(C) The number of nuclei remaining after time $t$ is given by $N = N_0 \left( \frac{1}{2} ight)^n$,where $n = \frac{t}{T_{1/2}}$.Given $N_0 = 4 	imes 10^{16}$,$T_{1/2} = 10 \ days$,and $t = 30 \ days$.First,calculate the number of half-lives: $n = \frac{30}{10} = 3$.The number of nuclei remaining is $N = 4 	imes 10^{16} 	imes \left( \frac{1}{2} ight)^3 = 4 	imes 10^{16} 	imes \frac{1}{8} = 0.5 	imes 10^{16}$.The number of nuclei that have decayed is $N_{decayed} = N_0 - N$.$N_{decayed} = 4 	imes 10^{16} - 0.5 	imes 10^{16} = 3.5 	imes 10^{16}$.

$A$ radioactive element initially has $4 \times 10^{16}$ active nuclei. If its half-life is $10 \ days$,find the number of nuclei that have decayed in $30 \ days$.

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