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1800-102-2727Let me ask you one question. What is your age? You will definitely answer this question as you are aware of the date when you were born. But suppose if in your locality you find this giant tree which has fallen due to a cyclone and you want to know the age of this tree. How would you be able to know the age of the fallen tree? So, a famous scientist W.S. Libby along with his colleagues developed a technique to determine the age of the living organism which died with the help of C-14 content present in the sample and was named the carbon dating technique. Let me tell you one interesting fact about W.S. Libby. He was awarded noble prize in the year 1960 as his development revolutionized archaeology and palaeontology. Lets have a tour of this article to understand more about the radiocarbon dating technique.
Table of Contents
It was introduced by W.S. Libby to find the age of fossils, old wood etc with the help of C-14.Radiocarbon dating is used to determine the age of carbon-containing substances. The C-14 technique is based on the following assumption:
As in the case of living material ratio of C-14 to C-12 remains relatively constant but when tissue in an animal or plant dies, C-14 decreases as there is no consumption and utilisation. Therefore the ratio of C-14 to C-12 decreases depending upon the age of the tissue. The decay constant is determined by using the half-life period of C-14.
A sample of dead tissue is converted to carbon dioxide and then analysed for the ratio of C-14 to C-12.
We know that the radioactive disintegration equation can be represented as,
${\Rightarrow \lambda =\frac{2.303}{t}}_{l}og\frac{{N}_{o}}{N}....\left(i\right)$
Here,
‘’ is called a decay constant or disintegration constant
‘t’ represents the time required for the disintegration.
‘No' represents the activity of the initial radioactive sample per unit mass.
‘N' represents the activity of disintegrated radioactive sample per unit mass.
We know that in case of living tissue or living material C-14 to C-12 remains relatively constant and comparing the same ratio for wood in equation (i) we get,
${\Rightarrow t=\frac{2.303}{\lambda}}_{l}og\frac{{N}_{o}}{N}$
Here,
‘t’ represent the age of the sample
‘No' represents the ratio $\frac{C-14}{C-12}$ in living tissue or living material or activity of living material per unit mass
'N' represents the ratio $\frac{C-14}{C-12}$ in wood or activity of wood per unit mass
Decay constant will be calculated by using the half-life of C-14,
${(t}_{1/2}{)}_{C-14}=\frac{ln2}{\lambda}=\frac{0.693}{\lambda}.....\left(ii\right)$
Here '' is the decay constant of C-14 isotope.
Q. A piece of wood was found to have $\frac{C-14}{C-12}$ ratio 0.5 times that in a living plant. Calculate the age of the dead plant. Given: half-life of C-14 isotope is 5760 years.
Answer: 5775 years
Solution: According to the given data,
Ratio of $\frac{C-14}{C-12}$ in wood sample = 0.5 (Ratio of $\frac{C-14}{C-12}$ in living green plants)
$\Rightarrow \frac{Ratioof\frac{C-14}{C-12}inlivinggreenplants\left({N}_{o}\right)}{Ratioof\frac{C-14}{C-12}inwoodsample\left(N\right)}=2$
Half life of C-14 isotope = 5760 years
Let the age of the dead plant be 't' years.
Decay constant will be calculated by using the half-life of C-14,
${(t}_{1/2}{)}_{C-14}=\frac{ln2}{\lambda}=\frac{0.693}{\lambda}$
Here '' is the decay constant of C-14 isotope.
Putting the value of half-life of C-14 we get,
$\lambda =\frac{ln2}{5760}\simeq 0.00012yea{r}^{-1}$
We know that radioactive disintegration equation can be represented as,
${\Rightarrow t=\frac{2.303}{\lambda}}_{l}og\frac{{N}_{o}}{N}$
Putting the value of decay constant () and value of (No N) in above equation, we get;
$\Rightarrow t=\frac{2.303}{0.00012}log2$
t5775 years
Q. The amount of C-14 present in the piece of wood is found to be 14 times of that is present in a fresh piece of wood. Calculate the age of the wood. (Given: half-life of C-14 is 5577 years).
Answer: 11,182 years
Solution: According to the given data,
Ratio of $\frac{C-14}{C-12}$ in wood sample = 0.25(Ratio of $\frac{C-14}{C-12}$ in fresh piece wood)
$\Rightarrow \frac{Ratioof\frac{C-14}{C-12}infreshwoodsample\left({N}_{o}\right)}{Ratioof\frac{C-14}{C-12}inwoodsample\left(N\right)}=4$
Half life of C-14 isotope = 5577 years
Let the age of the wood be 't' years
Decay constant will be calculated by using the half-life of C-14,
${(t}_{1/2}{)}_{C-14}=\frac{ln2}{\lambda}=\frac{0.693}{\lambda}$
Here '' is the decay constant of C-14 isotope
Putting the value of half-life of C-14 in equation(ii) we get,
$\lambda =\frac{ln2}{5577}\simeq 0.000124yea{r}^{-1}$
We know that radioactive disintegration equation can be represented as,
${\Rightarrow t=\frac{2.303}{\lambda}}_{l}og\frac{{N}_{o}}{N}$
Putting the value of decay constant () and value of (No N) in above equation, we get;
${\Rightarrow t=\frac{2.303}{0.000124}}_{l}og4$
$t\simeq 11,182years$
Q. An archaeologist unearthen a bone sample and to determine the age of the bone they reached the chemist and found that 35 % of the initial amount of carbon-14 is present in bone sample if the half life of C-14 is 5730, the age of bone will be:
Answer:
Solution: According to the given data,
Ratio of $\frac{C-14}{C-12}$ in bone sample = 0.35(Ratio of $\frac{C-14}{C-12}$ in fresh bone sample)
$\Rightarrow \frac{Ratioof\frac{C-14}{C-12}infreshbonesample\left({N}_{o}\right)}{Ratioof\frac{C-14}{C-12}inbonesample\left(N\right)}=\frac{100}{35}$
Half life of C-14 isotope = 5730 years
Let the age of the bone be 't' years.
Decay constant will be calculated by using the half-life of C-14,
${(t}_{1/2}{)}_{C-14}=\frac{ln2}{\lambda}=\frac{0.693}{\lambda}$
Here '' is the decay constant of C-14 isotope
Putting the value of half-life of C-14 in equation(ii) we get,
$\lambda =\frac{ln2}{5730}\simeq 0.000120yea{r}^{-1}$
We know that radioactive disintegration equation can be represented as,
${\Rightarrow t=\frac{2.303}{\lambda}}_{l}og\frac{{N}_{o}}{N}$
Putting the value of decay constant () and value of (No N) in the above equation, we get;
${\Rightarrow t=\frac{2.303}{0.000120}}_{l}og\frac{100}{35}$
$t\simeq 8750years$
Q. Determination of the age of fossils can be done using which of the following techniques?
Answer: (C)
Solution: In the case of living material ratio of C-14 to C-12 remains relatively constant but when tissue in animal or plant dies, C-14 decreases as there is no consumption and utilisation. Therefore the ratio of C-14 to C-12 decreases depending upon the age of the tissue. The decay constant is determined by using the half-life period of C-14. So, the age of the fossil is determined by measuring the ratio of C-14 and C-12 using the carbon dating technique.
Q. What are the limitations of radiocarbon dating?
Answer: Radiocarbon analysis cannot be used to date inorganic materials, and the procedure can be prohibitively expensive. Because of the low quantities of carbon-14 in older samples, dating samples which are older than 40,000 years is extremely challenging and the sample of age over 60,000 years old can't be dated.
Q. What is the difference between C-14 and C-12 atom?
Answer: C-14 and C-12 atoms are the two different isotopes of carbon atoms. Isotope is defined as the element having the same atomic number and different mass number due to which chemical properties of the substances are identical but the physical properties are different.
Q. What is the advantage of using a carbon-dating technique?
Answer: Radiocarbon dating is a worldwide dating technique that may be used anywhere in the globe as long as there is organic material present. It can build chronologies for locations that previously lacked calendars and can date from the last 50,000 years to roughly 400 years ago.
Q. What is the working principle of carbon dating technique?
Answer: Radiocarbon dating is a method of determining the carbon content of discrete organic items. It was introduced by W.S. Libby and his colleagues to find the age of fossils, old wood etc with the help ofC-14. The concept of radiocarbon dating was based on the assumption that when an organism died, it was cut off from the carbon cycle, resulting in a time capsule with a continuously decreasing carbon-14 concentration.