17 The half-life of C-14 is 5760 years.
When buried in the ground, a piece
of wood had 1% C-14. Post-
conversion to charcoal, the
percentage of C-14 went down to
0.26%. Approximately, how long
had the wood been buried?
(Take log103.8 = 0.58)
ps: A.
11293.68 years
B.
91287 years
C.
13891.7 years
D.
120341 years
Answers
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The wood had been buried for A. 11293.68 years.
Explanation:
t1/2 of C-14 = 5760 years
λ = 0.693/5760
∴ λ = 0.0001203 (year)⁻¹
From question, a piece of wood had 1% C-14
⇒ C¹⁴₀ = 1%
NC¹⁴ = 0.26%
The time taken to decay is given by the formula:
t = 2.303/λ log (N₀/N)
On substituting the values, we get,
t = 2.303/0.0001203 log (1/0.26)
t = 19143.8 × 0.585
∴ t = 11199.1 years
From the given option, 11293.68 years is closer to 11199.1 years.
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