Question

An ancient club is found that contains 100 g of pure carbon and has an activity of 6 decays per second. Determine its age assuming that in living trees the ratio of ¹⁴C/¹²C atoms is about 1.10 × 10^-12. Note that the half life of carbon-14 is 5700 years and the Avogadro number is 6.02 × 10²³.

Answer 1

Using the equation

N = N0*e^(-lambda*t)

t = (1/lambda)*ln (N0/N)

wealso know that

T_1/2 = ln 2/lambda

lambda = ln 2/T_1/2

lambda = (ln 2)/5700 = 1.21*10^-4 yr^-1

1 yr = 365*24*3600 sec

lambda = 3.83*10^-12 sec^-1

N0 = 1.1*10^-12*(0.1 kg)*1 amu/(1.67*10^-27 kg)*1atom/12 amu

N0 = 5.49*10^12 atoms

given that

|dN/dt| = 6 decay/sec

N = n0*e^(-lambda*t)

|dN/dt| = -N0*lambda*e^(-lambda*t)

|dN/dt| = -lambda*N

N = (dN/dt)/lambda

N = 6/(3.83*10^-12) = 1.567*10^12

So,

t =(1/lambda)*ln (N0/N)

t = (1/(3.83*10^-12))*ln ((5.49*10^12)/(1.567*10^12))

t = 3.273*10^11 sec = 10371.73 yrs

So the club is 10371.73 yrs old.