Question

The half-life for the radioactive decay of carbon-14 to nitrogen-14 is 5.73 x 10^3 years. Suppose nuclear chemical analysis shows that there is 0.523mmol of nitrogen-14 for every 1.000 mmol of carbon-14 in a certain sample of rock.
Calculate the age of the rock. Round your answer to 2 significant digits.

Answer 1

Answer : The time passed in years is $6.28\times 10^3\text{ years}$

Explanation :

Half-life = $5.73\times 10^3\text{ years}$

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

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

$k=\frac{0.693}{5.73\times 10^3\text{ years}}$

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

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  = 1.000 mol

a - x = amount left after decay process = 1.000 - 0.532 = 0.468 mol

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

$t=\frac{2.303}{1.21\times 10^{-4}}\log\frac{1.000}{0.468}$

$t=6276.2\text{ years}=6.28\times 10^3\text{ years}$

Therefore, the time passed in years is $6.28\times 10^3\text{ years}$