A radioactive sample is undergoing α decay. At any time t₁, its activity is A and another time t₂, the activity is A/5. What is the average life time for the sample?

Solution

Physics sir · 4 steps

Activity decays exactly like the number of nuclei, so the ratio of two activities hands you λ directly.

1. Activity decays exponentially. Activity is A = λ N, and N falls as e^(-λ t). So activity follows the same law: A(t) = A₀ e^(-λ t).
2. Take the ratio. Divide the activity at t₁ by the activity at t₂. The starting activity A₀ cancels, and the ratio is A ÷ A/5 = 5.

   A(t₁)/A(t₂) = e^(λ (t₂ - t₁)) = 5 ⇒ λ = (ln 5)/(t₂ - t₁)

3. Mean life. The average life is the reciprocal of the decay constant, τ = 1/λ. Flipping the fraction gives τ = (t₂ - t₁)/(ln 5).
4. So. The average life time is (t₂ - t₁)/(ln 5), which is option (d).

The idea to keep. For any two readings of activity or of N, ln(ratio) = λ × (time gap). Mean life is 1/λ, and half-life is ln 2/λ. Do not mix the two up.

Answer from the NTA final answer key. Solution by PrepWiser.