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.

Answer 1

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