JEE Main 2022 · 26 July · morning · Single correct · +4 / −1

The disintegration rate of a certain radioactive sample at any instant is 4250 disintegrations per minute. 10 minutes later, the rate becomes 2250 disintegrations per minute. The approximate decay constant is :
(Take log ₁₀ 1.88=0.274)

  1. (a) 0.02 min ⁻¹
  2. (b) 2.7 min ⁻¹
  3. (c) 0.063 min ⁻¹
  4. (d) 6.3 min ⁻¹
Answer and solution

Answer (c) 0.063 min ⁻¹

Solution

Physics sir · 4 steps

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

  1. Activity decays exponentially. The rate R = λ N is proportional to N, so it follows the same law: R = R₀ e^(-λ t). Taking logs and switching to base 10 gives the working formula.

    λ = 2.303/t log₁₀ R₀/R

  2. Find the ratio. R₀/R = 4250/2250 = 1.89 ≈ 1.88, so log₁₀ R₀/R = 0.274.
  3. Put in numbers. With t = 10 min: λ = 2.303/10 × 0.274 = 0.2303 × 0.274 = 0.0631 min⁻¹.
  4. So. The decay constant is about 0.063 min⁻¹, which is option (c).

The idea to keep. Activity, number of nuclei and mass all fall off with the same e^(-λ t). Any ratio of them at two times gives λ = 2.303/t log₁₀(ratio).

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

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Asked about 110 times in JEE Main since 2020. This question is from 2022.