1 answer

An archaeologist graduate student found a leg bone of a large animal during the building of...

Question:

An archaeologist graduate student found a leg bone of a large animal during the building of a new science building . The bone had a carbon-14 decay rate of 14.8 disintegration per minute per gram of carbon .Living organisms have a decay rate of 15.3 disintegrations per minute .How old is the bone? (a) 53.3 years. (b), 25 years. (c), 111years. (d), 83 years. (e), 275 years.


Answers

Radioactive decay follows first-order kinetics. By measuring the C-14 activity of a dead sample, its age can be calculated using the expression for radioactive decay as:

t = 2.303/) * log(No/N)

where t is its age, \lambda is the decay constant of C-14, N_{0} is the C-14 activity in a fresh sample of the species, N_{t} is its C-14 activity at present.

Given, N_{t} = 14.8 dpm/g (dpm/g is disintegration per minute per gram)

N_{0} = 15.3 dpm/g

Half-life of carbon-14, t_{1/2} = 5760 years

A = 0.693t12 = 0.693 / (5760 years) = 1.203*10-4 years-1

Substituting these values in the equation t = 2.303/) * log(No/N) ,

Age  t = [ 2.303/(1.203*10-4 years-1) ] * log (15.3 dpm/g / 14.8 dpm/g)

= [ 2.303/(1.203*10-4) ] * log (15.3/14.8) years

= 19143.807 * 0.0144 years

= 275.67 years

Hence the bone is 275 years old.

The correct option is option e - 275 years.

.

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