Question

The half life of $C - 14$ is $5600$ years. A sample of freshly cut wood from a tree contains ${\text{10mg}}$ of $C - 14$. The amount left in the sample after $50000$ years is ${\text{(a - x)}} \times {\text{100}}$ . The value of ${\text{(a - x)}} \times {\text{100}}$ is:
A.${\text{1mg}}$
B.${\text{2mg}}$
C.${\text{3mg}}$
D.${\text{4mg}}$

Answer 1

Hint: In a chemical reaction, the time taken by the species to reduce itself to half of its initial value is known as half life time. The half life of a species depends on their decomposition, if it is a zero order reaction, half life will be inversely proportional to initial concentration and if it is a first order reaction then its half life is independent of initial concentration of the species.

Complete step by step answer:
The half life of a species basically tells us the time taken by it to reduce itself to half of its initial concentration.
As we have given that the half life of $C - 14$ is $5600$ years.
And we know that this decomposition is a first order reaction, so our half life time is independent of initial concentration.
And we know that
${\text{K = }}\dfrac{{0.693}}{{{{\text{t}}_{\dfrac{1}{2}}}}}$ ---------------(1)
Where,
${\text{K}}$is the first order rate constant
${{\text{t}}_{\dfrac{1}{2}}}$ is the half life time.
And we have given that ${{\text{t}}_{\dfrac{1}{2}}}$ is $5600$ years.
So by putting the value in eq(1)
${\text{K = }}\dfrac{{0.693}}{{5600}}$ ----------------(2)
Now, for first order reaction
${\text{t = }}\dfrac{{2.303}}{{\text{K}}}{\text{log}}\dfrac{{\text{a}}}{{{\text{a - x}}}}$ ----------------(3)
And we have provided that after $50000$ years the amount left is ${\text{(a - x)}} \times {\text{100}}$ -----------(4)
So, using eq(2), eq(3) and eq(4)
${\text{50000 = }}\dfrac{{2.303 \times 5600}}{{0.693}}{\text{log}}\dfrac{{\text{a}}}{{{\text{a - x}}}}$
${\text{log}}\dfrac{{\text{a}}}{{{\text{a - x}}}} = \dfrac{{50000 \times 0.693}}{{2.303 \times 5600}}$
Here ${\text{a}}$ is the initial concentration which is provided to us, i.e. ${\text{10mg}}$.
${\text{log}}\dfrac{{10}}{{{\text{a - x}}}} = \dfrac{{50000 \times 0.693}}{{2.303 \times 5600}}$
By solving we get,
$ \Rightarrow {\text{a - x = 0}}{\text{.02mg}}$
$ \Rightarrow ({\text{a - x)}} \times {\text{100 = 0}}{\text{.02}} \times {\text{100 = 2mg}}$
Hence option (B) is correct.

Note:
As we are discussing in our problem the rate constant ${\text{K}}$ , basically it is the proportionality constant relating the rate of the reaction with concentration of the reactants. It is determined experimentally.