Given:
The percentage of C-14 in the sample was 73% found in living beings.
Half-life of C-14, t1/2 = 5730yrs = 1.81 * 10¹¹s
Decay constant of the C - 14 isotope
λ = ln(2) / t1/2
λ = In(2) / 1.81 * 10¹¹
λ = 3.83 * 10⁻¹²/s
Let the N₀ be the number of C- 14 in a living sample. Then the number of C-14 in the given sample will be,
N = 0.73 * N₀
Let t be the age of the fossil.
then the given number of leftover atoms of carbon isotope can be expressed in the form of
N = N₀e⁻λt
−λt = In(0.73)
So the age of the fossil,
t = In(0.73)/ −λ
t = In(0.73)/ -3.38 * 10⁻¹²
t = 8.22 * 10¹⁰ s
t = 2605.59 yrs
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