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The age of the bone is 11459 years
We'll begin by calculating the rate constant.
Half-life (t½) = 5730 years
[tex]K = \frac{0.693}{t1/2} \\ \\ K \: = \frac{0.963}{5730} \\ \\ K = 1.21 \times {10}^{ - 4} /year[/tex]
Original activity (N₀) = 15.2 cpm
Remain activity (N) = 3.80 cpm
Rate constant (K) = 1.21×10¯⁴ / year
[tex]log( \frac{N₀}{N}) = \frac{Kt}{2.303} \\ \\ log( \frac{15.2}{3.8}) = \frac{1.21 \times {10}^{ - 4} \times t}{2.303} \\ \\ log(4) = \frac{1.21 \times {10}^{ - 4} \times t}{2.303} \\ \\ cross \: multiply \\ \\ 2.303 \times log4 = 1.21 \times {10}^{ - 4} \times t \\ \\ divide \: both \: side \: by \: 1.21 \times {10}^{ - 4} \\ \\ t = \frac{2.303 \times log4}{1.21 \times {10}^{ - 4}} \\ \\ t \: = 11459 \: years[/tex]
Therefore, the bone is 11459 years old.
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