If the half-life of a radioactive nucleus is | vedclass

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(D) Given, half-life $t_{1/2} = 3$ days.Number of active nuclei remaining after time $t$ is given by $N(t) = N_0 (1/2)^{t/t_{1/2}}$.At the start of th

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$0.13$

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(D) Given, half-life $t_{1/2} = 3$ days.Number of active nuclei remaining after time $t$ is given by $N(t) = N_0 (1/2)^{t/t_{1/2}}$.At the start of the third day ($t = 2$ days), the number of nuclei remaining is $N_1 = N_0 (1/2)^{2/3} = N_0 / (2^{2/3}) = N_0 / (4^{1/3})$.Given $\sqrt[3]{0.25} \approx 0.63$, we have $N_1 = N_0 	imes 0.63$.At the end of the third day ($t = 3$ days), the number of nuclei remaining is $N_2 = N_0 (1/2)^{3/3} = 0.5 N_0$.The number of nuclei that decay during the third day is the difference between the nuclei present at the start and end of the day:$\Delta N = N_1 - N_2 = 0.63 N_0 - 0.5 N_0 = 0.13 N_0$.Thus, the fraction of the initial nuclei that decay on the third day is $0.13$.

If the half-life of a radioactive nucleus is $3$ days, nearly what fraction of the initial number of nuclei will decay on the third day? (Given, $\sqrt[3]{0.25} \approx 0.63$)

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