Question

The ratio of 14C to 12C in living organisms is 1.3 × 10-12. The fossilized remains of an organism are discovered and the ratio of 14C to 12C in the fossil is measured to be 4.6 × 10-13. How long ago, in years was the organism alive? (The half life of 14C is 5,730 years.)

Answer 1

Answer:

t = 8588 years

Step-by-step explanation:

From the given information:

Using the formula:

$k = \dfrac{In (2)}{t_{1/2}}$

Given that:

$t_{1/2}$ = 5730 years

Then:

$k = \dfrac{In (2)}{5730}$

$k = \dfrac{0.6931472}{5730}$

k = 1.20968 × 10⁻⁴

However; the expression to find the value of x is:

$x= x_oe^{-kt}$

$4.6 \times 10^{-13} = 1.3 \times 10^{-12} \times e^{(-1.20968\times 10^{-4} \times t)}$

$\dfrac{4.6 \times 10^{-13} }{1.3 \times 10^{-12}}= e^{(-1.20968\times 10^{-4} \times t)}$

$0.353846= e^{(-1.20968\times 10^{-4} \times t)}$

(-1.20968 × 10⁻⁴ × t) = In(0.353846)

(-1.20968 × 10⁻⁴ × t) = -1.03889

$t = \dfrac{-1.03889}{-1.20968 \times 10^{-4}}$

t = 8588 years