Question:

Using a nuclear counter the count rate of emitted particles from a radioactive source is measured. At t = 0 it was 1600 counts per second and t = 8 seconds it was 100 counts per second. The count rate observed, as counts per second, at t = 6 seconds is close to -

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The rate falls from 1600 to 100, a factor of 16 = \(2^4\), in 8 s, so the half-life is 2 s.
Updated On: Nov 5, 2026
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The Correct Option is A

Solution and Explanation

\(\frac{1600}{100} = 16 = 2^4\), so 4 half-lives pass in 8 s: \(T_{1/2} = 2\) s.
At \(t=6\) s, \(n = \frac{6}{2} = 3\) half-lives have passed.
Rate \(= \frac{1600}{2^3} = 200\) counts per second.
Answer: (1) 200
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