Question #3f7bf

2 Answers
Jul 24, 2017

Answer:

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 #(C-14)/(C-12) = 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 #(C-14)/(C-12)# in the sample has fallen to #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:

#(1/2)^x = 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

Answer:

(c) 22,920 years

Explanation:

If the ratio of #(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 #N/N_0 =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.