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(B) The activity of a radioactive sample at time $t$ is given by $A(t) = A_0 e^{-\lambda t}$.Given that at $t = 1 	ext{ hour}$,$A(1) = \frac{A_0}{\sq

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$\frac{A_0}{9}$

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(B) The activity of a radioactive sample at time $t$ is given by $A(t) = A_0 e^{-\lambda t}$.Given that at $t = 1 	ext{ hour}$,$A(1) = \frac{A_0}{\sqrt{3}}$.Substituting this into the formula: $\frac{A_0}{\sqrt{3}} = A_0 e^{-\lambda(1)}$,which implies $e^{-\lambda} = \frac{1}{\sqrt{3}}$.We need to find the activity after $3$ hours more,which means at a total time $t = 1 + 3 = 4 	ext{ hours}$.$A(4) = A_0 e^{-\lambda(4)} = A_0 (e^{-\lambda})^4$.Substituting $e^{-\lambda} = \frac{1}{\sqrt{3}}$ into the equation:$A(4) = A_0 \left(\frac{1}{\sqrt{3}}ight)^4 = A_0 \left(\frac{1}{3^2}ight) = \frac{A_0}{9}$.

The activity of a radioactive sample reduces from $A_0$ to $\frac{A_0}{\sqrt{3}}$ in $1$ hour. What will be the activity after $3$ hours more?

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