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 disintegrations per minute per gram of carbon. Living organisms have a decay rate of 15.3 disintegrations per minute. How old is the bone?

Answer 1

Answer : The time passed in years is $2.74\times 10^2\text{ years}$

Explanation :

Half-life of carbon-14 = 5730 years

First we have to calculate the rate constant, we use the formula :

$k=\frac{0.693}{t_{1/2}}$

$k=\frac{0.693}{5730\text{ years}}$

$k=1.21\times 10^{-4}\text{ years}^{-1}$

Now we have to calculate the time passed.

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant  = $1.21\times 10^{-4}\text{ years}^{-1}$

t = time passed by the sample  = ?

a = initial amount of the reactant disintegrate = 15.3

a - x = amount left after decay process = 14.8

Now put all the given values in above equation, we get

$t=\frac{2.303}{1.21\times 10^{-4}}\log\frac{15.3}{14.8}$

$t=274.64\text{ years}=2.74\times 10^2\text{ years}$

Therefore, the time passed in years is $2.74\times 10^2\text{ years}$