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(A) Let $N_0$ be the initial number of nuclei and $N$ be the number of undecayed nuclei after time $t$.Given that $99\%$ of the nuclei have decayed,th

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$6	au_{1/2}$ and $7	au_{1/2}$

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(A) Let $N_0$ be the initial number of nuclei and $N$ be the number of undecayed nuclei after time $t$.Given that $99\%$ of the nuclei have decayed,the number of undecayed nuclei is $N = N_0 - 0.99N_0 = 0.01N_0$.Using the radioactive decay law: $N = N_0(1/2)^n$,where $n = t/	au_{1/2}$ is the number of half-lives.Substituting the values: $0.01N_0 = N_0(1/2)^n \Rightarrow (1/2)^n = 0.01 \Rightarrow 2^n = 100$.Taking the logarithm on both sides: $n \log_{10}(2) = \log_{10}(100)$.$n 	imes 0.3010 = 2$.$n = 2 / 0.3010 \approx 6.64$.Since $n = t/	au_{1/2}$,we have $t = n 	imes 	au_{1/2} \approx 6.64 	au_{1/2}$.This value lies between $6	au_{1/2}$ and $7	au_{1/2}$.

$99\%$ of the nuclei of a radioactive element decay between which of the following times?

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