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Question
A sample of old wood shows 7.0 dps/g. If the fresh sample of tree shows 16.0 dps/g, how old is the given sample of wood? (Half-life of 14C is 5730 y)
Solution
Given: t1/2 of 14C = 5730 y
`(- "dN"_0)/"dt" = 16`dps/g
`(- "dN")/"dt"` = 7 dps/g
To find: Age of sample of wood
Formulae:
- `"t"_(1//2) = 0.693/lambda`
- `lambda = 2.303/"t" log_10 ("N"_0/"N")`
Calculation:
`lambda = 0.693/"t"_(1//2) = 0.693/5730 = 1.209 xx 10^-4 "y"^-1`
`"t" = 2.303/lambda log_10 ("N"_0/"N")`
`= 2.303/(1.209 xx 10^-4) log_10 (16.0/7.0)`
= 6839 y
The age of sample of old wood is 6839 y.
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