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Carbon-14 has a half-life of 5730 yr. A living organism has an activity of 15.2 counts per minute (cpm) per gram of carbon. If a bone is determined to have an activity of 3.80 cpm per gram of carbon, how old is the bone

Sagot :

Eduard22sly

The age of the bone is 11459 years

We'll begin by calculating the rate constant.

Half-life (t½) = 5730 years

Rate constant (K) =?

[tex]K = \frac{0.693}{t1/2} \\ \\ K \: = \frac{0.963}{5730} \\ \\ K = 1.21 \times {10}^{ - 4} /year[/tex]

  • Finally, we shall determine the age of the bone.

Original activity (N₀) = 15.2 cpm

Remain activity (N) = 3.80 cpm

Rate constant (K) = 1.21×10¯⁴ / year

Time (t) =?

[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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