Question

After discovering an ancient buried artifact made of wood, you decide to perform a radiometric analysis to determine the age of the artifact. You know that half of the mass of wood is in the form of Carbon. You also know that about one out of every trillion (10¹²) Carbon atoms in living plant material is a Carbon-14 atom (the vast majority of Carbon atoms are Carbon-12). If you analyze a 0.21-kg sample of your wooden artifact, there would be 5.268×10¹²  Carbon-14 atoms. The decay constant for carbon-14 is 3.833×10^-12 s^(-1).

1. What was the activity (or decay rate) of the sample when the tree was first cut down?

2. When you take your sample into the laboratory you find that it has an activity (or decay rate) of 10.51 Bq. How long ago was the tree from which this wood came cut down? Assuming the artifact was made right after the tree was cut down, this result should tell you the age of the artifact. Give your answer in years (yr)

3. What will be the activity (or decay rate) of this sample 3500 years in the future?

Answer 1

Q1.

eqution for radioactive decay:

N=N0*exp(-lambda*t)

where N0=original number of atoms

labda=decay constant

N=number of undecayed atoms at time t

given that lambda=3.833*10^(-12)

original mass of carbon=0.21 kg/2=0.105 kg

let number of atoms of C14=x

then number of atoms of C12=10^12*x

total mass=((x/(6.023*10^23))*14)+((10^12*x/(6.023*10^23))*12) grams

=1.9924*10^(-11)*x

==>1.9924*10^(-11)*x=105

==>x=5.27*10^12

so initial activity=x*decay constant

=20.2 Bq

part 2:

activity at present time=10.51 Bq

then N(t)=activity/lambda

=2.742*10^12

using it in decay equation:

2.742*10^12=5.27*10^12*exp(-lambda*t)

==>t=1.7046*10^11 seconds

=5405.2 years

Q3.

number of undecayed nucei in 3500 years in future=3500+5405.2 years from beginning

=8905.2 years

will be given by

N0*exp(-lambda*time)

=5.27*10^12*exp(-3.833*10^(-12)*8905.2*365*24*3600)

=1.7961*10^12

so activity=1.7961*10^12*3.833*10^(-12)

=6.8845 Bq