Activities of three radioactive substances A, B and C are represented by the curves A, B and C, in the figure. Then their half-lives T_(1/2)(A) : T_(1/2)(B) : T_(1/2)(C) are in the ratio:

Solution

Physics sir · 3 steps

On a ln R versus t graph, the slope magnitude is λ, so read the three slopes and flip them.

1. Straight-line form. Activity falls as R = R₀ e^(-λ t). Taking the log gives a straight line whose slope is -λ.

   ln R = ln R₀ - λ t, T_(1/2) = (ln 2)/λ

   Curve	Drop in ln R	Time taken	λ (per year)
   A	6 → 0, so 6	10 yr	6/10 = 0.6
   B	6 → 0, so 6	5 yr	6/5 = 1.2
   C	2 → 0, so 2	5 yr	2/5 = 0.4

2. Invert the slopes. Since T_(1/2) ∝ 1/λ, the ratio is 1/0.6 : 1/1.2 : 1/0.4. Multiply each term by 1.2 to get 2 : 1 : 3.
3. So. T_(1/2)(A) : T_(1/2)(B) : T_(1/2)(C) = 2 : 1 : 3, which is option (d). C has the gentlest slope, so it has the longest half-life.

The idea to keep. On a ln R versus t plot, only the slope matters: steeper line means larger λ and shorter half-life. Where the line starts or ends tells you nothing alone.

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