Question

Activity of a radioactive sample decreases to \(\left(\frac{1}{3}\right)^{rd}\) of its original value in \(3\) days. Then in \(9\) days its activity will become

• A. \(\left(\frac{1}{27}\right)\) of the original value
• B. \(\left(\frac{1}{9}\right)\) of the original value
• C. \(\left(\frac{1}{18}\right)\) of the original value
• D. \(\left(\frac{1}{3}\right)\) of the original value

Hint:

Activity falls by the same factor in equal time intervals, so nine days is three intervals of three days.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
Activity follows \(A=A_0e^{-\lambda t}\), so it decreases by a fixed factor in equal time intervals.

Step 2: Use the given data
In 3 days the activity becomes \(\dfrac{1}{3}\) of the original value, so \(e^{-3\lambda}=\dfrac{1}{3}\).

Step 3: Calculation
In 9 days, \(A=A_0e^{-9\lambda}=A_0\left(e^{-3\lambda}\right)^3=A_0\left(\dfrac{1}{3}\right)^3=\dfrac{A_0}{27}\).

Step 4: Compare the options
This is option (A).

Final Answer:
After 9 days the activity is one twenty-seventh of the original value. \[ \boxed{\text{(A) }\dfrac{1}{27}\text{ of the original value}} \]