The activity of a radioactive sample falls from 700 s –1 to 500 s –1 in 30 minutes. Its half life is close to:

Solution

Physics sir · 4 steps

Activity decays exponentially just like the number of nuclei, so take a log of the ratio 700/500.

1. Activity follows decay law. Activity is A = λ N, so it falls exactly like N: A = A₀ e^(-λ t). Taking the log gives λ t = ln A₀/A.

   λ = 1/t ln A₀/A = ln(700/500)/(30 min) = (ln 1.4)/(30 min)

2. Put in numbers. ln 1.4 ≈ 0.336, so λ ≈ 0.0112 min⁻¹.
3. Half-life. T_(1/2) = (ln 2)/λ = (0.693 × 30)/0.336 ≈ 61.8 min.
4. So. The half-life is close to 62 min, option (a).

The idea to keep. Activity is proportional to the number of nuclei left, so use T_(1/2) = (t ln 2)/ln(A₀/A) directly. Never assume the drop is linear in time.

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