KCET · Physics · Nuclear Physics
A radioactive sample has 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
- A 10 years
- B 7 years
- C 15 years
- D 5 years
Answer & Solution
Correct Answer
(B) 7 years
Step-by-step Solution
Detailed explanation
Given, \(T_{1 / 2}=3\) years
We know that
\(R=R_0 e^{-\lambda t}\)
Here, \(R=\frac{R_0}{5}\)
\(\begin{aligned} & \Rightarrow \quad \frac{R_0}{5}=R_0 e^{-\lambda t} \Rightarrow \frac{1}{5}=e^{-\lambda t} \\ & \Rightarrow \ln \left(9^{-1}=\ln \left(e^{-\lambda t}\right)\right. \\ & \Rightarrow \quad t=\frac{\ln 5}{\lambda}=\frac{\ln 5}{\frac{0.693}{t_{1 / 2}}}=\frac{\ln 5}{0.693} \times t_{1 / 2}=\frac{\ln 5}{0.693} \times 3 \\ & \quad=6.96 \approx 7 \text { years }\end{aligned}\)
We know that
\(R=R_0 e^{-\lambda t}\)
Here, \(R=\frac{R_0}{5}\)
\(\begin{aligned} & \Rightarrow \quad \frac{R_0}{5}=R_0 e^{-\lambda t} \Rightarrow \frac{1}{5}=e^{-\lambda t} \\ & \Rightarrow \ln \left(9^{-1}=\ln \left(e^{-\lambda t}\right)\right. \\ & \Rightarrow \quad t=\frac{\ln 5}{\lambda}=\frac{\ln 5}{\frac{0.693}{t_{1 / 2}}}=\frac{\ln 5}{0.693} \times t_{1 / 2}=\frac{\ln 5}{0.693} \times 3 \\ & \quad=6.96 \approx 7 \text { years }\end{aligned}\)
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