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.