The activity of a radioactive sample reduces | vedclass

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(B) The activity of a radioactive sample follows the law $A(t) = A_0 e^{-\lambda t}$.Given that at $t_1 = 1 	ext{ hour}$,$A(t_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 $t_1 = 1 	ext{ hour}$,$A(t_1) = \frac{A_0}{\sqrt{3}}$.Substituting this into the decay formula: $\frac{A_0}{\sqrt{3}} = A_0 e^{-\lambda(1)} \implies e^{-\lambda} = \frac{1}{\sqrt{3}} = (3)^{-1/2}$.We need to find the activity after $3 	ext{ hours}$ more,which means at a total time $t_2 = 1 + 3 = 4 	ext{ hours}$.$A(t_2) = A_0 e^{-\lambda(4)} = A_0 (e^{-\lambda})^4$.Substituting $e^{-\lambda} = (3)^{-1/2}$:$A(4) = A_0 ((3)^{-1/2})^4 = A_0 (3)^{-2} = \frac{A_0}{9}$.

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

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