In a sample of ^{66}Cu,7/8 of the initial num | vedclass

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(A) The number of nuclei remaining after time $t$ is given by $N = N_0 \left( \frac{1}{2} \right)^{\frac{t}{T_{1/2}}}$.Given that $7/8$ of the nuclei

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$5$

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(A) The number of nuclei remaining after time $t$ is given by $N = N_0 \left( \frac{1}{2} \right)^{\frac{t}{T_{1/2}}}$.Given that $7/8$ of the nuclei have decayed,the remaining fraction is $1 - 7/8 = 1/8$.So,$\frac{N}{N_0} = \frac{1}{8}$.Substituting this into the formula: $\frac{1}{8} = \left( \frac{1}{2} \right)^{\frac{15}{T_{1/2}}}$.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,$T_{1/2} = \frac{15}{3} = 5 \ min$.

In a sample of $^{66}Cu$,$7/8$ of the initial number of nuclei decay into $Zn$ in $15 \ min$. The half-life of the sample is ......... $min$.

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