Question

A wooden artifact from a Chinese temple has a 14C activity of 42.8 counts per minute as compared with an activity of 58.2 counts per minute for a standard of zero age. Part A From the half-life for 14C decay, 5715 yr, determine the age of the artifact.

Answer 1

Answer : The age of the artifact is, $2.54\times 10^3\text{ years}$

Explanation :

Half-life = 5715 years

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

$k=\frac{0.693}{5715\text{ years}}$

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

Now we have to calculate the time taken to decay.

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

t = time taken by sample = ?

a = initial activity of the reactant  = 58.2 counts per minute

a - x = activity left after decay process  = 42.8 counts per minute

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

$t=\frac{2.303}{1.21\times 10^{-4}}\log\frac{58.2}{42.8}$

$t=2540.5\text{ years}=2.54\times 10^3\text{ years}$

Therefore, the age of the artifact is, $2.54\times 10^3\text{ years}$