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Question
65% of 111In sample decays in 4.2 d. What is its half-life?
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Solution
Given: N0 = 100, N = 100 - 65 = 35, t = 4.2 d
To find: t1/2
Formulae:
- `lambda = 2.303/"t" log_10 ("N"_0/"N")`
- `"t"_(1//2) = 0.693/lambda`
Calculation:
1. `lambda = 2.303/"t" log_10 ("N"_0/"N")`
`lambda = 2.303/4.2 log_10 (100/35)`
= 0.548 × 0.456 = 0.2499 d-1
2. `"t"_(1//2) = 0.693/lambda`
`= 0.693/0.2499 = 2.773 "d"`
Half-life of 111In sample is 2.773 d.
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