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(B) For a radioactive substance,the amount remaining after $n$ half-lives is given by $N = N_0(1/2)^n$. Given that $75\%$ disintegrates,the amount rem

Vedclass · Accepted Answer

$30$

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(B) For a radioactive substance,the amount remaining after $n$ half-lives is given by $N = N_0(1/2)^n$. Given that $75\%$ disintegrates,the amount remaining is $25\%$ of the initial amount,i.e.,$N = 0.25 N_0$. So,$0.25 N_0 = N_0(1/2)^n$,which simplifies to $(1/2)^2 = (1/2)^n$. Thus,$n = 2$ half-lives. Since $2$ half-lives correspond to $60 \ min$,the half-life $t_{1/2} = 60 \ min / 2 = 30 \ min$.

Half-life of a radioactive substance which disintegrates by $75\%$ in $60 \ min$ will be ......... $min$.

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