The half life period of radioactive substance is 60 days. The time taken for 7th8 of its original mass to disintegrate will be

78 disintegrate means 18 remains∴ NN0=12n⇒18=12n∴n=3n=tT1/2 ∴t=nT12 =3×60=180 days

The half life period of radioactive substance is 60 days. The time taken for 7th8 of its original mass to disintegrate will be

78 disintegrate means 18 remains∴ NN0=12n⇒18=12n∴n=3n=tT1/2 ∴t=nT12 =3×60=180 days

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

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$\frac{7}{8}$ disintegrate means $\frac{1}{8}$ remains

$∴$ $\frac{N}{N_{0}} = {(\frac{1}{2})}^{n} \Rightarrow \frac{1}{8} = {(\frac{1}{2})}^{n}$ $∴ n = 3$

$n = \frac{t}{T_{1 / 2}}$ $∴ t = n T_{\frac{1}{2}}$ $= 3 \times 60 = 180$ days

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The half life period of radioactive substance is 60 days. The time taken for 7th8 of its original mass to disintegrate will be

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