Question

An archeologist finds an ancient fire pit containing partially consumed firewood, and the carbon-14 content of the wood is only 10.4 % that of an equal carbon sample from a present-day tree. What is the age in years of the ancient site? Your answer should be in the form of N x 10⁴ years. Enter only the number N with two decimal places, do not enter unit.

Carbon-14 has a half-life of 5 730 years

Answer 1

given

T1/2 = 5730 years

decay constant, lamda = 0.693/T1/2

= 0.693/5730

= 1.209*10^-4 year^-1

let t years is the age of the ancient site.

let No is initial number of atom of C-14

N is the number of C-14 atoms present after t years.

N = 10.4% of No

N = 0.104*No

we know, N = No*e^(-lamda*t)

0.104*No = No*e^(-lamda*t)

0.104 = e^(-lamda*t)

ln(0.104) = -lamda*t

==> t = -ln(0.104)/lamda

= -ln(0.104)/(1.209*10^-4)

= 1.87*10^4 years <<<<<<<<-----------------Answer