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Question
A 3/4 of the original amount of radioisotope decays in 60 minutes. What is its half-life?
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Solution
Given: N0 = 100,
For N, `100 xx 3/4 = 75` ∴ 100 - 75 = 25, t = 60 min
To find: t1/2
Formulae:
- `lambda = 2.303/"t" log_10 ("N"_0/"N")`
- `"t"_(1//2) = 0.693/lambda`
Calculation:
- `lambda = 2.303/"t" log_10 ("N"_0/"N")`
`= 2.303/60 log_10 (100/25)`
`= 2.303/60 log_10 4`
= 0.0231 min-1 - `"t"_(1//2) = 0.693/lambda = 0.693/0.0231` = 30 min
Half-life of the radioisotope is 30 min.
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