Posted by Leila on .
An ancient wooden club is found that contains 270g of carbon and has an activity of 8.3 decays per second.
Determine its age assuming that in living trees the ratio of 14C/12C atoms is about 1.3x10^12?
t= in years

Physics Need help please 
Elena,
The decay constant is
λ=ln2/T = 0.693/5730 = 1.209•10⁻⁴ yr⁻¹.
The fraction of ¹⁴C << ¹²C, => we use the atomic weight of ¹²C to find the number of carbon atoms in 270 g:
N= (270/12)•6.022•10²³ =1.35•10²⁵ atoms,
The number of ¹⁴C in the sample of living tree is
N₁₄=1.3•10⁻¹²•1.35•10²⁵ =1.76•10¹³ nuclei.
Decay Law
N=N₀•exp(λt)
Activity
A= λN,
λN=λN₀•exp(λt) (1 year= 3.16•10⁷ s),
8.3 = (1.209•10⁻⁴•1.76•10¹³/3.16•10⁷) exp(1.209•10⁻⁴•t),
exp(1.209•10⁻⁴•t) =
=8.3•3.16•10⁷/1.209•10⁻⁴•1.76•10¹³)=0.12,
1.209•10⁻⁴•t = ln0.12 =  2.095,
t=2.095/1.209•10⁻⁴ = 1.73•10⁴ yr.