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Question

(B) The activity of a radioactive sample follows the law $A(t) = A_0 e^{-\lambda t}$.Given that at $t = 1 	ext{ hour}$,$A(1) = \frac{A_0}{\sqrt{3}}$.

Vedclass · Accepted Answer

$\frac{A_0}{9}$

Vedclass · Answer

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

The activity of a sample reduces from $A_0$ to $A_0 / \sqrt{3}$ in one hour. The activity after $3$ hours more will be

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