Starting with a sample of pure ^{66}Cu,frac{7 | vedclass

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(A) Given that $\frac{7}{8}$ of the sample decays,the remaining fraction of the sample is $1 - \frac{7}{8} = \frac{1}{8}$.The law of radioactive decay

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$5$

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(A) Given that $\frac{7}{8}$ of the sample decays,the remaining fraction of the sample is $1 - \frac{7}{8} = \frac{1}{8}$.The law of radioactive decay is given by $N = N_0 \left( \frac{1}{2} \right)^{\frac{t}{T_{1/2}}}$,where $N$ is the remaining amount,$N_0$ is the initial amount,$t$ is the time elapsed,and $T_{1/2}$ is the half-life.Substituting the values: $\frac{1}{8} = \left( \frac{1}{2} \right)^{\frac{15}{T_{1/2}}}$.Since $\frac{1}{8} = \left( \frac{1}{2} \right)^3$,we have $\left( \frac{1}{2} \right)^3 = \left( \frac{1}{2} \right)^{\frac{15}{T_{1/2}}}$.Equating the exponents: $3 = \frac{15}{T_{1/2}}$.Therefore,$T_{1/2} = \frac{15}{3} = 5 \ min$.

Starting with a sample of pure $^{66}Cu$,$\frac{7}{8}$ of it decays into $Zn$ in $15 \ min$. The corresponding half-life is .......... $min$.

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