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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: - Fusion reaction takes place at high temperature because
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Topics: Atoms and Nuclei
(136)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Fusion reaction takes place at high temperature because
- A planar coil having 12 turns carries 15 A current. The coil is oriented with respect
- A Si and a Ge diode has identical physical dimensions. The bandgap in Si is larger
- If a gymnast, sitting on a rotating stool with his arms outstretched, suddenly
- In a parallel plate capacitor of capacitance C, a metal sheet is inserted
Topics: Atoms and Nuclei (136)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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