The half-life period of a radioactive substan | vedclass

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(A) The number of half-lives $n$ in a total time $t$ is given by $n = \frac{t}{T_{1/2}}$.Given $t = 20\, min$ and $T_{1/2} = 5\, min$,we have $n = \fr

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$93.75$

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(A) The number of half-lives $n$ in a total time $t$ is given by $n = \frac{t}{T_{1/2}}$.Given $t = 20\, min$ and $T_{1/2} = 5\, min$,we have $n = \frac{20}{5} = 4$.The fraction of the substance remaining after $n$ half-lives is given by $N/N_0 = (1/2)^n$.Substituting $n = 4$,we get $N/N_0 = (1/2)^4 = 1/16$.The fraction of the substance that has decayed is $1 - N/N_0 = 1 - 1/16 = 15/16$.To express this as a percentage,we calculate $(15/16) 	imes 100\% = 93.75\%$.

The half-life period of a radioactive substance is $5\, min$. The amount of substance decayed in $20\, min$ will be..........$\%$

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