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Question
0.5 g sample of 201Tl decays to 0.0788 g in 8 days. What is its half-life?
Solution
Given: N0 = 0.5 g, N = 0.0788 g, t = 8 days
To find: t1/2
Formulae:
- `lambda = 2.303/"t" log_10 ("N"_0/"N")`
- t1/2 = `0.693/lambda`
Calculation:
1. `lambda = 2.303/"t" log_10 ("N"_0/"N")`
`= 2.303/8 log_10 (0.5/0.0788)`
`= 2.303/8 log_10 (6.3452)`
`= 2.303/8 xx 0.8024`
λ = 0.231 d-1
2. `"t"_(1//2) = 0.693/lambda`
`"t"_(1//2) = 0.693/0.231`
`"t"_(1//2)` = 3d
The half-life of 201Tl is 3d.
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