Posted by Sandhya on .
A specimen taken from the wrappings of a mummy contains 7.02 g of carbon and has an activity of 1.34 Bq. How old is the mummy? Determine its age in years assuming that in living trees the ratio of 14C/12C atoms is 1.23E12
I am not able to get the answer.Please help

physicsRadioactivity 
DrBob222,
Do you have an answer?

physicsRadioactivity 
Sandhya,
Yes. But it is incorrect
R = 1.34 Bq = 1.34 decays / second Mass of sample = 7.02 g
For 14 C T 1/2 = 5700 years (approximate value)
= 5700 x 365 x 24 x 60 x 60 =1.8 x 10 11 seconds
T 1/2 = 0.6931 / λ
λ = 0.6931 / 1.8 x 10 11 = 3.86 x 10 12 per second
Number of atoms contained in 7.02 g of carbon (12 C atoms)
= (7.02 x 6.023 x 10 23) /14 = 3.02 x 10 23
14 C atoms / 12 C atoms = 1.23 x 1012
No of 14 C atoms (radioactive) = No of 12 C atoms x 1.23 x 1012
N0 = 3.02 x 10 23 x 1.23 x 1012 = 3.7146 x 10 11
R = Nλ
N = R/λ = 1.34 / 3.86 x 10 12 = 3.4715 x 10 11
t = 3.323 x 5700 x In (3.4715 x 10 11 / 3.7146 x 10 11 ) = 557 years