Question

At time \(t=0\), the activity of a radioactive substance is \(1600\) Bq. At \(t=8\) s, the activity remains \(100\) Bq. The activity at \(t=2\) s is

• A. \(400\) Bq
• B. \(800\) Bq
• C. \(1600\) Bq
• D. \(3200\) Bq

Hint:

Find the half-life from the fall in activity between 0 and 8 s, then see how many half-lives pass in 2 s.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Activity halves in every half-life: \(A=A_0\left(\dfrac{1}{2}\right)^{t/T}\).

Step 2: Find the half-life
Activity falls from 1600 Bq to 100 Bq, a factor of 16, which is \(2^4\). So 4 half-lives pass in 8 s, giving \(T=\dfrac{8}{4}=2\) s.

Step 3: Calculation
At \(t=2\) s exactly one half-life has passed, so \(A=\dfrac{1600}{2}=800\) Bq.

Final Answer:
The activity at 2 s is 800 Bq, which is option (B). \[ \boxed{\text{(B) }800\text{ Bq}} \]