If 87.5 % of atoms of a radioactive element d | vedclass

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(D) The number of atoms remaining after time $t$ is given by $N(t) = N_0 (1/2)^{t/T}$, where $T$ is the half-life.Given that $87.5 \%$ of atoms decay

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$\frac{15}{16}$

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(D) The number of atoms remaining after time $t$ is given by $N(t) = N_0 (1/2)^{t/T}$, where $T$ is the half-life.Given that $87.5 \%$ of atoms decay in $6 \ days$, the remaining fraction is $100 \% - 87.5 \% = 12.5 \% = 1/8$.So, $1/8 = (1/2)^{6/T}$.Since $1/8 = (1/2)^3$, we have $6/T = 3$, which gives $T = 2 \ days$.Now, we need to find the fraction of atoms that decay in $8 \ days$.The fraction remaining after $8 \ days$ is $N(8)/N_0 = (1/2)^{8/T} = (1/2)^{8/2} = (1/2)^4 = 1/16$.The fraction of atoms that decay is $1 - (	ext{fraction remaining}) = 1 - 1/16 = 15/16$.

If $87.5 \%$ of atoms of a radioactive element decay in $6 \ days$, then the fraction of atoms of the element that decay in $8 \ days$ is

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