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Solve · Q36

Q.65% of 111In sample decays in 4.2 d. What is its half life?
(Ans. : 2.77 d)

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Step 1. If 65% of the sample decays in 4.2 days, then 35% remains: N/N0 = 0.35.

Step 2. t1/2 = t x ln(2) / ln(N0/N) = 4.2 x 0.6931 / ln(1/0.35) = 2.9111 / ln(2.857) = 2.9111 / 1.0498 ≈ 2.773 d. …

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