A radioactive sample S 1 having an activity of 5 μCi has twice the number of nuclei as another sample S 2 which has an activity of 10 μCi. The half lives of S 1 and S 2 can be

Activity of S 1 = 1 2 (activity of S 2 ) or λ 1 N 1 = 1 2 λ 2 N 2 Or λ 1 λ 2 = N 2 2 N 1 T 1 T 2 = 2 N 1 N 2 or (T = half - life = ℓ n 2 λ ) Given N 1 = 2 N 2 , Therefore T 1 T 2 = 4

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PhysicsModern PhysicsRadioactivityEasy2 minPYQ_2020

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A radioactive sample S₁having an activity of 5 μCi has twice the number of nuclei as another sample S₂which has an activity of 10 μCi. The half lives of S₁and S₂can be

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Answer:

Solution:

Activity of S₁= $\frac{1}{2}$ (activity of S₂)

or $λ_{1} N_{1} = \frac{1}{2} (λ_{2} N_{2})$

Or $\frac{λ_{1}}{λ_{2}} = \frac{N_{2}}{2 N_{1}} \frac{T_{1}}{T_{2}} = \frac{2 N_{1}}{N_{2}}$ or (T = half - life = $\frac{ℓ n 2}{λ}$ )

Given N₁= 2 N₂,

Therefore $\frac{T_{1}}{T_{2}} = 4$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:Modern PhysicsSubtopic:Radioactivity

⏱ 2mℹ️ Source: PYQ_2020

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