
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
487.5k+ views
Hint: 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.
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.
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