An artifact classified as charcoal, found in a site at Longmoor, UK, is found to

Question

An artifact classified as charcoal, found in a site at Longmoor, UK, is found to have a 14C radioactivity of 8.72×10-2 counts per second per gram of carbon. If living carbon-containing objects have an activity of 0.255 counts per second per gram of carbon, estimate the age of the artifact? The half-life of 14C is 5730 years An artifact classified as bone, horse, found in a site at Belloy-sur-Somme, France, is found to have a 14C radioactivity of 7.34×10-2 counts per second per gram of carbon. If living carbon-containing objects have an activity of 0.255 counts per second per gram of carbon, estimate the age of the artifact? The half-life of 14C is 5730 years

Explanation / Answer

1. Given,

Activity living carbon-containing object (Ao) = 0.255 counts per second per gram of carbon

Activity of artifact as charcoal found (A) = 8.72x10-2 counts per second per gram of carbon

We know the relation between the activity and no. of molecules. That is

Ao/A = No/N = 0.255/8.72x10-2 = 2.92 ............................(1)

The radioactive decay follows first order kinetics. So,

t= (2.303/k) log(No/N) ...........................(2)

Here, No = initial no. of atoms

N = final no. of atoms

t = life time

k = rate constant = 0.693/t1/2 = 0.693/5730 = 1.22 x 10-4 yr-1 [Given t1/2 of 14C = 5730 years]

Putting (1) and above k value in (2), we get

t = (2.303/1.22 x 10-4 yr-1) log(2.92) = 8797.15 yr

Hence, the age of artifact as charcoal is 8797.15 yr.

2. Given,

Activity living carbon-containing object (Ao) = 0.255 counts per second per gram of carbon

Activity of artifact as bone of horse found (A) = 7.34x10-2 counts per second per gram of carbon

We know the relation between the activity and no. of molecules. That is

Ao/A = No/N = 0.255/7.34x10-2 = 3.47 ............................(1)

The radioactive decay follows first order kinetics. So,

t= (2.303/k) log(No/N) ...........................(2)

Here, No = initial no. of atoms

N = final no. of atoms

t = life time

k = rate constant = 0.693/t1/2 = 0.693/5730 = 1.22 x 10-4 yr-1 [Given t1/2 of 14C = 5730 years]  

Putting (1) and above k value in (2), we get

t = (2.303/1.22 x 10-4 yr-1) log(3.47) = 10209.54 yr

Hence, the age of artifact as bone of horse is 10209.54 yr.

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