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(D) The law of radioactive decay is given by $N = N_0 \left(\frac{1}{2}\right)^n$,where $n$ is the number of half-lives.Given that the material reduce

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$256$

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(D) The law of radioactive decay is given by $N = N_0 \left(\frac{1}{2}ight)^n$,where $n$ is the number of half-lives.Given that the material reduces to $1/8$ of its original amount in $3$ days:$\frac{N_0}{8} = N_0 \left(\frac{1}{2}ight)^n \implies \left(\frac{1}{2}ight)^3 = \left(\frac{1}{2}ight)^n \implies n = 3$.Since $3$ half-lives correspond to $3$ days,the half-life $T_{1/2} = 1$ day.After $5$ days,the number of half-lives elapsed is $n = \frac{5 	ext{ days}}{1 	ext{ day}} = 5$.The remaining amount is $N = 8 	imes 10^{-3} \, kg = 8 \, g$.Using the decay formula: $8 = N_0 \left(\frac{1}{2}ight)^5$.$8 = N_0 \left(\frac{1}{32}ight)$.$N_0 = 8 	imes 32 = 256 \, g$.

$A$ radioactive material is reduced to $1/8$ of its original amount in $3$ days. If $8 \times 10^{-3} \, kg$ of the material is left after $5$ days,the initial amount of the material is $....... \, g$.

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