Question

Enter your answer in the provided box. The usefulness of radiocarbon dating is limited to objects no older than 50,000 yr. What percent of the carbon−14, originally present in a 7.0−g sample, remains after this period of time? The half-life of carbon−14 is 5.73 × 103 yr.

Answer 1

Answer : The percent of the carbon−14 left is, 0.242 %

Explanation :

This is a type of radioactive decay and all radioactive decays follow first order kinetics.

To calculate the rate constant, we use the formula :

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

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

$k=1.205\times 10^{-4}\text{ years}^{-1}$

Now we have to calculate the amount left.

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

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

where,

k = rate constant  = $1.205\times 10^{-4}\text{ years}^{-1}$

t = time taken for decay process  = 50000 years

a = initial amount or moles of the reactant  = 7 g

a - x = amount or moles left after decay process  = ?

Putting values in above equation, we get:

$1.205\times 10^{-4}=\frac{2.303}{50000\text{ years}}\log\frac{7g}{a-x}$

$a-x=0.0169g$

The amount left of carbon-14 = 0.0169 g

Now we have to calculate the percent of the carbon−14 left.

$\text{Percent of carbon}-14\text{ left}=\frac{\text{Amount left of carbon}-14}{\text{Original amount of carbon}-14}\times 100$

$\text{Percent of carbon}-14\text{ left}=\frac{0.0169g}{7g}\times 100=0.242\%$

Therefore, the percent of the carbon−14 left is, 0.242 %