Question

A wooden piece is $11520$ years old. What is the fraction of $^{14}C$ activity left in the piece?
A. $0.12$
B. $0.25$
C. $0.50$
D. $0.75$

Answer 1

Hint: The age of the wooden piece has been given which can give us the value of half-life of the carbon isotope i.e. $^{14}C$. The fraction of carbon isotope which is left in the wooden sample is dependent on the activity of the carbon isotope and one can relate the half-life value to age to find an approach to answer.

Complete step by step answer:
-First of all, we will learn about the carbon activity concept in which activated carbon is also called activated charcoal. It is a form of carbon that is processed to have small-sized and low volume having pores that increase the overall surface area which is made available for adsorption or chemical reactions.
-Now, as the wooden piece is $11520$ years old we can take that value as time. The value of the half-life for the $^{14}C$ is \[5760\] years which is a known and universal value. Let's calculate the activity by putting the values in the activity equation.
$k = \dfrac{{2.303}}{t}\log \left( {\dfrac{{{N_0}}}{N}} \right)$
As we know the value of $k$ can to be taken as $k = \dfrac{{0.693}}{{{t_{1/2}}}}$ and let's put this value in the above equation we get as follow,
$\dfrac{{0.693}}{{{t_{1/2}}}} = \dfrac{{2.303}}{t}\log \left( {\dfrac{{{N_0}}}{N}} \right)$
Now put the values of half-life and time in the equation,
$\dfrac{{0.693}}{{5760}} = \dfrac{{2.303}}{{11520}}\log \left( {\dfrac{{{N_0}}}{N}} \right)$
Taking the value of to one side,
$\dfrac{N}{{{N_0}}} = \dfrac{{0.693 \times 11520}}{{5760 \times 2.303}} = 0.25$
-Hence, the value of $^{14}C$ activity left in the piece is $0.25$ which shows option B as a correct choice.
So, the correct answer is “Option B”.

Note: The activated carbon is generally gotten from charcoal. The carbon isotope $^{14}C$ and other naturally occurring radioactive substances which are present in our bodies contribute to the background radiation we receive. If a living tissue has $250$ decays per second in a kilogram then it gives $0.25$ decays per second for a gram of carbon which is present in that living tissue.