The half-life period of a radioactive substan | vedclass

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(C) Given that the half-life $T_{1/2} = 60 \ days$.If $\frac{7}{8}$ of the original mass disintegrates,the remaining mass $N$ is $N = N_0 - \frac{7}{8

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$180$

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(C) Given that the half-life $T_{1/2} = 60 \ days$.If $\frac{7}{8}$ of the original mass disintegrates,the remaining mass $N$ is $N = N_0 - \frac{7}{8}N_0 = \frac{1}{8}N_0$.The relation between remaining mass and initial mass is given by $N = N_0 \left(\frac{1}{2}ight)^n$,where $n$ is the number of half-lives.Substituting the values: $\frac{1}{8}N_0 = N_0 \left(\frac{1}{2}ight)^n$.This simplifies to $\left(\frac{1}{2}ight)^3 = \left(\frac{1}{2}ight)^n$,which gives $n = 3$.The total time taken $t = n 	imes T_{1/2} = 3 	imes 60 \ days = 180 \ days$.

The half-life period of a radioactive substance is $60 \ days$. The time taken for $\frac{7}{8}$ of its original mass to disintegrate will be $...... \ days$.

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