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.

Answer 1

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:

$t = (2.303/\lambda )*log (N_{0}/N_{t})$

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

$\lambda = 0.693/t_{1/2}$ = 0.693 / (5760 years) = 1.203*10^-4 years^-1

Substituting these values in the equation $t = (2.303/\lambda )*log (N_{0}/N_{t})$ ,

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.