- Carbon-14(C14) decays at a constant rate in such a way that it reduces to 50% in 5568 years. Find the age of an old wooden piece in which the carbon is only 25% of the original.
Asked by rajubarman | 1st Dec, 2019, 10:03: AM
λ= 0.693 /t1/2
t1/2 = 5568 years ... (given)
λ= 0.693 /5568 = 1.24 × 10-4 / year
By law of radioactive decay,
N/N0 = e-λt
The number of active nuclei at t = 0, N0 = 100%
And N = 25%
25/100 = e-λt
ln(0.25) = -λt
ln(0.25) = -1.24 × 10-4 × t
t = -1.3862/ (-1.24 × 10-4 ) ≈ 11180 years
Answered by Shiwani Sawant | 3rd Dec, 2019, 06:22: PM
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