Question

The half-life of ${C^{14}}$ is $5730$ year. What fraction of its original ${C^{14}}$ would left year $22920$ year of storage?
A. $0.50$
B. $0.25$
C. $0.125$
D. $0.0625$

Answer 1

To answer this question, we should know about the formulae of half-lives. According to the formula, we need to put the value of half –life which is already given in the question, then we get the value of the decay constant ( $k$ ). After this, we can calculate the value of $x$ which indicates the left fraction.
Formula used:
${t_{1/2}} = \dfrac{{0.693}}{k}$
$k = \dfrac{{2.303}}{t}\log \dfrac{1}{{1 - x}}$

Complete answer:
The half-life of the reaction is the time required for the reactant concentration to decrease to one half its initial value.
According to the question, it is given that the half-life of ${C^{14}}$ is $5730$ year.
In the given formula,
${t_{1/2}} = \dfrac{{0.693}}{k}$
We can write it as,
$k = \dfrac{{0.693}}{{{t_{1/2}}}}$
$k = \dfrac{{0.693}}{{5730}}$
Removing the decimal point,
$k = \dfrac{{693}}{{5730\, \times \,1000}}$
By solving this we get,
$k = 1.21\, \times \,{10^{ - 4\,}}\,yr{s^{ - 1}}$
Now, we know that,
$k = \dfrac{{2.303}}{t}\log \dfrac{1}{{1 - x}}$
Put the value of $k = 1.21\, \times \,{10^{ - 4\,}}\,yr{s^{ - 1}}$
Here, $t\, = 22900\,yrs$
$1.21\, \times \,{10^{ - 4\,}}\, = \dfrac{{2.303}}{{22900}}\log \dfrac{1}{{1 - x}}$
$\dfrac{{1.21\, \times \,{{10}^{ - 4\,}}\, \times 22900}}{{2.303}} = \log \dfrac{1}{{1 - x}}$
Solving this question, we get,
$1 - x = \dfrac{1}{{16}}$
$x = 1 - \dfrac{1}{{16}}$
Take LCM on right hand side and solve
$x = \dfrac{{16 - 1}}{{16}}$
$x = \dfrac{{15}}{{16}}$
We get the value of x
$x = 0.0625$
The fraction of its original ${C^{14}}$ would left year $22920$ year of storage is $x = 0.0625$.

Note:
In this question, we hear the term “half-life” many times. We should remember the meaning of this term “half –life”. The half-life of the reaction is the time required for the reactant concentration to decrease to one half its initial value. Half-life is also known as biological half-life. We should remember the formula for the calculation of decay constant which is represented as $k$ and half-life of the species which is represented as ${t_{1/2}}$. By remembering these formulas, we can calculate the decay constant, half-life, and remaining year of degradation or remaining amount of the species.