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Question
The half-life of 35S is 87.8 d. What percentage of 35S sample remains after 180 d?
Solution
Given: t1/2 = 87.8 d, N0 = 100, t = 180 d
To find: % of 35S that remains after 180 days
Formulae:
- `lambda = 0.693/("t"_(1//2))`
- `lambda = 2.303/"t" log_10 ("N"_0/"N")`
Calculation:
- `lambda = 0.693/("t"_(1//2)) = 0.693/(87.8 "d") = 7.893 xx 10^-3 "d"^-1`
- Now, `lambda = 2.303/"t" log_10 ("N"_0/"N")`
`log_10 ("N"_0/"N") = (lambda"t")/2.303`
`= (7.893 xx 10^-3 xx 180)/2.303`
= 0.617
Taking antilog on both sides we get,
`"N"_0/"N" = 4.140`
N = `100/4.140`
= 24.155 ≈ 24.2%
Percentage of 35S that remains after 180 d is 24.2%.
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