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* 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=(0.693)/(t_(1/2))`

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

`k=1.21* 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=(2.303)/(k)\log(a)/(a-x)`

where,

k = rate constant =
`1.21* 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=(2.303)/(1.21* 10^(-4))\log(15.3)/(14.8)`

`t=274.64\text{ years}=2.74* 10^2\text{ years}`

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