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

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(D) Given that $\frac{7}{8}$ of the sample of ${}^{66}Cu$ decays in $15 \ minutes$.The fraction of the sample remaining undecayed is $N = 1 - \frac{7}

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(D) Given that $\frac{7}{8}$ of the sample of ${}^{66}Cu$ decays in $15 \ minutes$.The fraction of the sample remaining undecayed is $N = 1 - \frac{7}{8} = \frac{1}{8}$.We know that the remaining amount $N$ after $n$ half-lives is given by $N = \left(\frac{1}{2}\right)^n$.Since $\frac{1}{8} = \left(\frac{1}{2}\right)^3$,the number of half-lives $n$ is $3$.The relationship between time $t$,number of half-lives $n$,and half-life $T$ is $n = \frac{t}{T}$.Substituting the values,$3 = \frac{15}{T}$.Therefore,the half-life $T = \frac{15}{3} = 5 \ minutes$.

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

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