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(A) The radioactive decay follows the formula $N = N_0 	imes (1/2)^n$,where $n$ is the number of half-lives.For the first case: $1.25 = 10 	imes (1/

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$5$

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(A) The radioactive decay follows the formula $N = N_0 	imes (1/2)^n$,where $n$ is the number of half-lives.For the first case: $1.25 = 10 	imes (1/2)^n \implies (1/2)^n = 1.25/10 = 1/8 = (1/2)^3$.Thus,$n = 3$ half-lives correspond to $15 \ days$.Therefore,the half-life period $T_{1/2} = 15/3 = 5 \ days$.For the second case: $N_0 = 1 \ kg = 1000 \ g$ and $N = 500 \ g$.$500 = 1000 	imes (1/2)^n \implies (1/2)^n = 500/1000 = 1/2 = (1/2)^1$.Thus,$n = 1$ half-life is required.Time taken = $n 	imes T_{1/2} = 1 	imes 5 = 5 \ days$.

$10 \ g$ of a radioactive substance is reduced to $1.25 \ g$ after $15 \ days$. Its $1 \ kg$ mass will reduce to $500 \ g$ in how many days?

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