A radioactive sample has a half-life of 3 yea | vedclass

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(B) Given, half-life $T_{1/2} = 3$ years.We know that the activity $R$ at time $t$ is given by $R = R_0 e^{-\lambda t}$, where $\lambda = \frac{\ln 2}

Vedclass · Accepted Answer

$7$

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(B) Given, half-life $T_{1/2} = 3$ years.We know that the activity $R$ at time $t$ is given by $R = R_0 e^{-\lambda t}$, where $\lambda = \frac{\ln 2}{T_{1/2}}$.We want the time $t$ when $R = \frac{R_0}{5}$.Substituting this into the equation: $\frac{R_0}{5} = R_0 e^{-\lambda t} \Rightarrow \frac{1}{5} = e^{-\lambda t}$.Taking the natural logarithm on both sides: $\ln(1/5) = -\lambda t \Rightarrow \ln(5) = \lambda t$.Therefore, $t = \frac{\ln 5}{\lambda} = \frac{\ln 5}{\ln 2 / T_{1/2}} = \frac{\ln 5}{\ln 2} 	imes T_{1/2}$.Using $\ln 5 \approx 1.609$ and $\ln 2 \approx 0.693$:$t = \frac{1.609}{0.693} 	imes 3 \approx 2.3218 	imes 3 \approx 6.965$ years.Rounding to the nearest integer, $t \approx 7$ years.

$A$ radioactive sample has a half-life of $3$ years. The time required for the activity of the sample to reduce to $\frac{1}{5}$th of its initial value is about (in $years$)

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