# Question 3f7bf

Jul 24, 2017

c) 22920 years

#### Explanation:

I am not 100% on this solution, so someone more knowledgeable than I should definitely feel free to jump in and correct if and where necessary.

Assumption : The question doesn't make sense if $\frac{C - 14}{C - 12} = \frac{1}{15}$ as the natural ratio is much, much smaller. That would mean that the ratio would have increased in the sample - impossible. Therefore I have assumed that the question meant that the ratio $\frac{C - 14}{C - 12}$ in the sample has fallen to $\frac{1}{15}$ of the current ratio.

Given the above assumption we have this:
((C-14)/(C-12))_s = ((C-14)/(C-12))_c × 1/15

Where subscript s denotes the sample's ratio and c the current ratio.

The question then is: how many half lives are required to reduce the ratio by that factor. Mathematically that can be expressed like this:

${\left(\frac{1}{2}\right)}^{x} = \frac{1}{15}$ where x is the number of half lives.

⇒ log((1/2)^x) = log(1/15)

⇒ xlog(1/2) = log(1/15)

⇒ x = log(1//15)/log(1//2) = 3.907 half lives

The question provides the half life of carbon-14, so we multiply the number of half lives by the time for each half life to determine the total length of time:

 t = 3.907 × 5730 = 22,386 years

This is not exactly equal to either of the four options (not ideal!), but it is closest to option (c) 22,920 years.

Jul 25, 2017

(c) 22,920 years

#### Explanation:

If the ratio of $\frac{C - 14}{C - 12}$ has decreased to 1/15 of initial then the abundance of C-14 has decreased to 1/15 of its initial value. Hence $\frac{N}{N} _ 0 = \frac{1}{15}$ where N is the current abundance and ${N}_{0}$ is the initial abundance.

Then use this equation: N=N_0 e^(-λ t)

⇒ N/N_0 = e^(-λ t)

⇒ ln( N/N_0) = -λ t

⇒ t = ln( N/N_0) / (-λ)

⇒ t = ln( N_0/N) / (λ)

Remember that tthe decay constant, λ is given by:
λ = ln (2)/ t_½ 

Then the equation becomes:
⇒ t = (ln( N_0/N) × t_½) / (ln (2)

Substitute in the values:
⇒ t = (ln( 15/1) × 5730) / (ln (2)) = 22,386# years

Option (c) is closest to this value, so that is the solution.