87.5% decomposition of a radioactive substanc | vedclass

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(D) For a radioactive substance,the amount remaining after $n$ half-lives is given by $N = N_0 	imes (1/2)^n$.Given that $87.5\%$ has decomposed,the

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(D) For a radioactive substance,the amount remaining after $n$ half-lives is given by $N = N_0 	imes (1/2)^n$.Given that $87.5\%$ has decomposed,the amount remaining is $100\% - 87.5\% = 12.5\%$.So,$12.5 = 100 	imes (1/2)^n$,which simplifies to $(1/2)^3 = (1/2)^n$.Therefore,$n = 3$ half-lives.Since $3$ half-lives correspond to $3$ hours,the half-life $t_{1/2} = 3 	ext{ hours} / 3 = 1 	ext{ hour}$.

$87.5\%$ decomposition of a radioactive substance completes in $3$ hours. What is the half-life of that substance?

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