Question

The activity of a radioactive sample is measured as N₀ counts per minute at t = 0 and (N₀ / e) counts Per minute at t = 5 min. The time (in min) at which activity reduces to half its value is

(A) log e (2/5)

(B) 5 log₁₀2

(C) 5 log_e2

(D) log₁₀^(2/5)

Answer 1

The correct option is  (C) 5 log_e2.

Explanation:

We know fraction left after time t is (N / N₀) = (1 / 2)^{[t / {T(1/2)}]}

where T_(1/2) is half life time

Given : at t = 5 minute, (N / N₀) = (1 / e)

Half life time means substance reduced to half of initial value hence

required time is 5 log_e2.