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(C) Given:Total time,$T = 10$ yearsHalf-life,$T_{1/2} = 5$ yearsNumber of half-lives,$n = \frac{T}{T_{1/2}} = \frac{10}{5} = 2$Using the radioactive d

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$75$

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(C) Given:Total time,$T = 10$ yearsHalf-life,$T_{1/2} = 5$ yearsNumber of half-lives,$n = \frac{T}{T_{1/2}} = \frac{10}{5} = 2$Using the radioactive decay law,the fraction remaining is given by $\frac{N}{N_0} = \left(\frac{1}{2}ight)^n$Substituting the value of $n$,we get $\frac{N}{N_0} = \left(\frac{1}{2}ight)^2 = \frac{1}{4}$This means $\frac{1}{4}$ of the original substance remains after $10$ years.The fraction decayed is $1 - \frac{N}{N_0} = 1 - \frac{1}{4} = \frac{3}{4}$To express this as a percentage,we multiply by $100$: $\frac{3}{4} 	imes 100 = 75 \%$Therefore,$75 \%$ of the sample will decay in $10$ years.

$A$ radioactive sample has a half-life of $5$ years. The percentage of the fraction decayed in $10$ years will be (in $\%$)

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