Carbon-14 has a half life of 5730 years. If a shell is found and has 52% of its original Carbon-14, how old is it?

2 Answers
Aug 16, 2016

Answer:

Approximately #5406# years

Explanation:

#-5730 * log_2(52/100) = -5730 * log(52/100)/log(2) ~~ 5406# years

The proportion of #""^14C# remaining, compared with the original is modelled by the function:

#p(t) = 2^-t/(5730)#

where #t# is measured in years.

Given that #p(t) = 52/100#, we have:

#52/100 = 2^(-t/5730)#

Taking logs base #2# we have:

#log_2(52/100) = -t/5730#

So multiplying both sides by #-5730# and transposing we get the formula:

#t = -5730*log_2(52/100)#

Footnote

The complexity of radiocarbon dating comes from two factors:

  • The proportion of #""^14C# to #""^12C# is very small to start with.
  • The proportion has been dramatically affected by factors like the industrial revolution, which replaced a significant portion (one third? one quarter?) of the #CO_2# in the atmosphere with carbon dioxide generated from fossil fuels - which contain virtually no remaining #""^14C#. As a result the method needs calibration to provide accurate dates.
Aug 16, 2016

Answer:

#sf(5409color(white)(x)"yr")#

Explanation:

The expression for radioactive decay is:

#sf(N_t=N_0e^(-lambdat))#

#sf(N_0)# is the initial number of undecayed atoms

#sf(N_t)# is the number of undecayed atoms at time #sf(t)#

#sf(lambda)# is the decay constant

Taking natural logs of both sides:

#sf(lnN_t=lnN_0-lambdat)#

#:.##sf(ln(N_t/N_0)=-lambdat)#

We can get the value of #sf(lambda)# using the expression:

#sf(lambda=0.693/t_(1/2)=0.693/5730=1.209xx10^(-4)color(white)(x)a^(-1))#

Putting in the numbers:

#sf(ln(52/100)=-1.208xx10^(-4)xxt=-0.6539)#

#:.##sf(t=0.6539/(1.209xx10^(-4))=5409color(white)(x)yr)#