# A bone was found to be approximately 24000 years old. The current amount of carbon-14 in the bone was 5.000 atoms. How many atoms must the bone have had originaly, assuming that the half-life of carbon-14 is approximately 6000 years?

Nov 22, 2016

#### Explanation:

The equation for 1st order decay is:

sf(N_t=N_0e^(-lambdat)

Taking natural logs of both sides $\Rightarrow$

$\textsf{\ln {N}_{t} = \ln {N}_{0} - l a m \mathrm{da} t}$

$\textsf{\lambda}$ is the decay constant which is inversely related to the 1/2 life:

$\textsf{\lambda = \frac{0.693}{t} _ \left(\frac{1}{2}\right) = \frac{0.693}{6000} = 0.0001155 \textcolor{w h i t e}{x} {a}^{- 1}}$

We need to find $\textsf{{N}_{0}}$ so:

$\textsf{\ln \left(5\right) = \ln \left({N}_{0}\right) - 0.0001155 \times 24000}$

$\therefore$$\textsf{\ln \left({N}_{0}\right) = 1.609 + 2.772 = 4.381}$

From which:

$\textsf{{N}_{0} = 79.9}$

Nov 23, 2016