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The decay constant of a radio isotope is λ. If A₁ and A₂ are its activities at time t₁ and t₂ respectively, the number of nuclei which have decayed during the time (t₁ – t₂):
Options
(a) λ (A₁ – A₂)
(b) A₁t₁ -A₂t₂
(c) A₁ – A₂
(d) (A₁ – A₂) / λ
Correct Answer:
(A₁ – A₂) / λ
Explanation:
Activity is given by A = dN/ dt = -λN
Activity at time t₁ is A₁ = – λN₁
and activity at time t₂ is A₂ = -1 N₂ as t₁ > t₂ therefore, number of atoms remained after time t₁ is less than that remained after time t₂.
That is, N₁ < N₂. number of nuclei decayed in (t₁ - t₂)
= N₂ - N₁ = (A₁ - A₂) / λ
Related Questions:
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- Hydrogen atom in ground state is excited by a monochrmatic radiation of λ=975 Å
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Topics: Atoms and Nuclei
(136)
Subject: Physics
(2479)
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