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(B) The activity of a radioactive sample at time $t$ is given by $A = A_0 \left(\frac{1}{2}\right)^{t/T_{1/2}}$, where $T_{1/2}$ is the half-life.Alte

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$\frac{1}{81}$

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(B) The activity of a radioactive sample at time $t$ is given by $A = A_0 \left(\frac{1}{2}ight)^{t/T_{1/2}}$, where $T_{1/2}$ is the half-life.Alternatively, using the general decay law $A = A_0 \left(\frac{1}{n}ight)^{t/t_0}$, where the activity becomes $\frac{1}{n}$ of its initial value in time $t_0$.Given that the activity becomes $\frac{1}{3}$ of its initial value in $t_0 = 20 	ext{ hours}$, we have $\frac{A}{A_0} = \left(\frac{1}{3}ight)^{t/t_0}$.For $t = 80 	ext{ hours}$, the fraction remaining is $\frac{A}{A_0} = \left(\frac{1}{3}ight)^{80/20}$.$\frac{A}{A_0} = \left(\frac{1}{3}ight)^4 = \frac{1}{81}$.

An active nucleus decays to one-third $\left(\frac{1}{3}\right)$ of its initial activity in $20 \text{ hours}$. The fraction of original activity remaining after $80 \text{ hours}$ is:

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