Question:
If the activities of a radioactive substance at times \(t=0\) and \(t=3T\) are A and B respectively, then the activity of the substance at a time \(t=9T\) is
If the activities of a radioactive substance at times \(t=0\) and \(t=3T\) are A and B respectively, then the activity of the substance at a time \(t=9T\) is
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Whenever activity is given at two different times, form ratios first. Exponential terms cancel quickly and lengthy calculations are avoided.
Updated On: Jun 22, 2026
- \(\frac{A^{3}}{B^{2}}\)
- \(\frac{A^{2}}{B}\)
- \(\frac{B^{2}}{A}\)
- \(\frac{B^{3}}{A^{2}}\) \bigskip
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The Correct Option is D
Solution and Explanation
Concept:
Activity of a radioactive substance decreases exponentially with time.
\[
R=R_0e^{-\lambda t}
\]
where
\[
R_0=\text{initial activity}
\]
\[
\lambda=\text{decay constant}
\]
The ratio method is the fastest way to solve such questions.
Step 1: Write the activities at the given instants.
At \(t=0\), \[ A=A \] At \(t=3T\), \[ B=Ae^{-3\lambda T} \] Therefore, \[ e^{-3\lambda T} = \frac{B}{A} \]
Step 2: Find the activity at \(t=9T\).
\[ R=Ae^{-9\lambda T} \] But \[ e^{-9\lambda T} = \left(e^{-3\lambda T}\right)^3 \] Hence, \[ R = A \left( \frac{B}{A} \right)^3 \] \[ R = \frac{AB^3}{A^3} \] \[ R = \frac{B^3}{A^2} \]
Step 3: State the final answer.
Therefore the activity at \(t=9T\) is \[ \boxed{\frac{B^3}{A^2}} \]
Step 1: Write the activities at the given instants.
At \(t=0\), \[ A=A \] At \(t=3T\), \[ B=Ae^{-3\lambda T} \] Therefore, \[ e^{-3\lambda T} = \frac{B}{A} \]
Step 2: Find the activity at \(t=9T\).
\[ R=Ae^{-9\lambda T} \] But \[ e^{-9\lambda T} = \left(e^{-3\lambda T}\right)^3 \] Hence, \[ R = A \left( \frac{B}{A} \right)^3 \] \[ R = \frac{AB^3}{A^3} \] \[ R = \frac{B^3}{A^2} \]
Step 3: State the final answer.
Therefore the activity at \(t=9T\) is \[ \boxed{\frac{B^3}{A^2}} \]
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