Radioactivity
posted by Sandhya on .
1) An old wooden tool is found to contain only 11.9 percent of 14C that a sample of fresh wood would. How many years old is the tool?
2) 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.
Please check my answers
1)
For 14 C T 1/2 = 5570 years N/N0 = 11.9/100 = 0.119
N / N0 = e λ t 2.303 x In 0.119 = λt λ =0.6931 / 5570 =1.244 x 10 4
t = (2.303 x In0.119 x 5570) / 0.6931 = 17110 years
2)
R = 1.34 Bq = 1.34 decays / second
14 C atoms / 12 C atoms = 1.23 x 1012 = N0/N N/N0 = 10 12 /1.2
Number of atoms contained in 7.02 g of carbon
= (7.02 x 6.023 x 10 23) /14 = 3.02 x 10 23 N = 3.02 x 10 23
R = Nλ λ = 1.34 / 3.02 x 10 23 = 0.4437 x 10 23
2.303 x In (10 12 /1.23 ) = 0. 4437 x 10 23 x t
t = 61.82 x 10 23 seconds = 1.96 x 10 17 years

T 1/2 of C14=5730 yrs.
Here is the answer for the 1st one, hope it will be helpfull.
Lamda= ln2/(T1/2)
= 0.693/5730=1.209E4
N/N0= 11.9/100 = 0.119
N=N0 e^lamda*t
N/N0=e^lamda*t
ln(0.119)= 1.209E4t
t=ln(0.119)/(1.209E4)
t=17606.5 yrs. 
T 1/2 of C14=5730 yrs.
Here is the answer for the 1st one, hope it will be helpfull.
Lamda= ln2/(T1/2)
= 0.693/5730=1.209E4
N/N0= 11.9/100 = 0.119
N=N0 e^lamda*t
N/N0=e^lamda*t
ln(0.119)= 1.209E4t
t=ln(0.119)/(1.209E4)
t=17606.5 yrs.
(ps: this is also for my Montgomery College, Takoma Park college Dr. Shaleen Shukla PH 204 class)