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Question
Sample of carbon obtained from any living organism has a decay rate of 15.3 decays per gram per minute. A sample of carbon obtained from very old charcoal shows a disintegration rate of 12.3 disintegrations per gram per minute. Determine the age of the old sample given the decay constant of carbon to be 3.839 × 10−12per second.
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Solution
Data: 15.3 decays per gram per minute (living organism), 12.3 disintegrations per gram per minute (very old charcoal). Hence, we have,
`("A"("t"))/"A"_0 = 12.3/15.3`, λ = 3.839 x 10-12 per second
A(t) = A0e-λt ∴ `"e"^(lambda"t") = "A"_0/"A"`
∴ `lambda"t" = log_"e"("A"_0/"A")`
∴ t = `2.303/lambda log_10 ("A"_0/"A")`
`= 2.303/(3.839 xx 10^-12)log_10(15.3/12.3)`
`= (2.303 xx 10^12)/3.839`(log 15.3 - log 12.3)
`= (2.303 xx 10^12)/3.839 (1.1847 - 1.0899)`
`= ((2.303)(0.0948))/3.839 xx 10^12`s
= 5.687 x 1010 s
`= (5.687 xx 10^10 "s")/(3.156 xx 10^7 "s per year")`
= 1802 years
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