A radioactive substance decays to left(frac{1 | vedclass

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(A) The activity of a radioactive substance follows the relation $A = A_0 \left(\frac{1}{2}\right)^n$,where $n$ is the number of half-lives.Given that

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(A) The activity of a radioactive substance follows the relation $A = A_0 \left(\frac{1}{2}ight)^n$,where $n$ is the number of half-lives.Given that the activity decays to $\frac{1}{16}$ of its initial value,we have $\frac{1}{16} = \left(\frac{1}{2}ight)^n$.Since $\frac{1}{16} = \left(\frac{1}{2}ight)^4$,we find that $n = 4$.The total time taken is $T = n 	imes t_{1/2}$,where $t_{1/2}$ is the half-life.Given $T = 80\, days$,we have $80 = 4 	imes t_{1/2}$.Therefore,$t_{1/2} = \frac{80}{4} = 20\, days$.

$A$ radioactive substance decays to $\left(\frac{1}{16}\right)^{th}$ of its initial activity in $80\, days$. The half-life of the radioactive substance expressed in days is ... .

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