(D) The half-life $(T_{1/2})$ of the radioactive sample is $20 \ days$.After $n$ half-lives,the remaining amount of the substance is given by $N = N_0

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$7/8$ part of the substance disintegrates in $60 \ days$

(D) The half-life $(T_{1/2})$ of the radioactive sample is $20 \ days$.After $n$ half-lives,the remaining amount of the substance is given by $N = N_0(1/2)^n$.In $60 \ days$,the number of half-lives elapsed is $n = 60 / 20 = 3$.The remaining amount of the substance is $N = N_0(1/2)^3 = N_0 / 8$.The amount of substance that has disintegrated is $N_0 - N = N_0 - N_0/8 = 7N_0/8$.Therefore,$7/8$ part of the substance disintegrates in $60 \ days$.

Class 12 Physics Nuclei Law of Radioactivity by Rutherford and Soddy and Half Life and Mean Life

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The half-life of a radioactive sample is $20 \ days$. This means that:

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A
the substance completely disintegrates in $40 \ days$
B
the substance completely disintegrates in $80 \ days$
C
$1/8$ part of the substance disintegrates in $60 \ days$
D
$7/8$ part of the substance disintegrates in $60 \ days$

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Law of Radioactivity by Rutherford and Soddy and Half Life and Mean Life

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The half-life of a radioactive sample is $20 \ days$. This means that:

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