Question

Radiocarbon dating of a sample of wood from the tomb of an Egyptian pharaoh found that in the 1 gram of the sample there were 9.843 times 10^-15 mol of C-14 and 1.202 times 10^-2 mol of C-12. The C-14 isotope is radioactive and decays with a half-life of 5, 730 years: the C-12 isotope is stable (does not decay). If the initial ratio of these isotopes is known to have been Number of C-14/Number of C-12 = 1.20 times 10^-12, how old is the sample of wood in the tomb?

Answer 1

decay equation for C-14 is given by

N(t)=N0*exp(-k*t)

where N0=original amount

k=decay constant

at t=5730 years, N(t)=N0/2

==>1/2=exp(-k*5730)

==>k=1.21*10^(-4)

let the sampe is T years old

then after T years,

number of c-14/number of C-12=9.843*10^(-15)/(1.202*10^(-2))

=>original number of C-14*exp(-k*T)/original number of C-12=8.188*10^(-13)

=> 1.2*10^(-12)*exp(-1.21*10^(-4)*T)=8.188*10^(-13)

==>T=3158.12 years

so the sample is 3158.12 years old.