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MHT CET 2026 (19 May, Shift 2) · Law of Radioactive Decay

Law of Radioactive Decay question from MHT CET 2026 (19 May, Shift 2)

Q12MHT CET 2026 · 19 MAY · SHIFT 2Medium

Law of Radioactive Decay

In an alpha decay, the daughter nucleus produced is itself unstable and undergoes further decay. If the number of parent and daughter nuclei at time tt are NpN_p and NdN_d respectively and their decay constants are λp\lambda_p and λd\lambda_d respectively. What condition needs to be satisfied in order for NdN_d to remain constant?

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Answer: (d)

  1. Daughter is formed at rate λpNp\lambda_p N_p and decays at rate λdNd\lambda_d N_d.
  2. dNddt=λpNp−λdNd=0\frac{dN_d}{dt} = \lambda_p N_p - \lambda_d N_d = 0 for constant NdN_d.
  3. Npλp=NdλdN_p\lambda_p = N_d\lambda_d, so the answer is (d).
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