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(A) The radioactive decay follows first-order kinetics,where the activity $A$ at time $t$ is given by $A = A_0 e^{-\lambda t}$.Given that the initial

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(A) The radioactive decay follows first-order kinetics,where the activity $A$ at time $t$ is given by $A = A_0 e^{-\lambda t}$.Given that the initial activity $A_0 = 10A$,where $A$ is the permissible safe activity level.The decay constant $\lambda$ is related to the half-life $t_{1/2}$ by $\lambda = \frac{\ln 2}{t_{1/2}} = \frac{0.693}{30 	ext{ days}}$.Using the first-order integrated rate equation: $t = \frac{2.303}{\lambda} \log_{10} \left( \frac{A_0}{A} ight)$.Substituting the values: $t = \frac{2.303}{(0.693 / 30)} \log_{10} (10)$.Since $\frac{2.303}{0.693} \approx \frac{1}{0.434} \approx 3.32$ (or more precisely $\frac{2.303}{0.693} \approx 3.32$ and $3.32 	imes 30 \approx 100$),we get $t = 30 	imes \frac{2.303}{0.693} 	imes 1 = 30 	imes 3.325 \approx 100 	ext{ days}$.

$A$ radioactive element gets spilled over the floor of a room. Its half-life period is $30$ days. If the initial activity is ten times the permissible value,after how many days will it be safe to enter the room?

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