Chemical Kinetics - Result Question 50
Passage Based Questions
Passage
Carbon-14 is used to determine the age of organic material. The procedure is based on the formation of ${ }^{14} Cby$ neutron capture in the upper atmosphere.
$ { } _7^{14} N+{ } _0 n^{1} \longrightarrow{ } _6^{14} C+{ } _1 p^{1} $
${ }^{14} C$ is absorbed by living organisms during photosynthesis. The ${ }^{14} C$ content is constant in living organism once the plant or animal dies, the uptake of carbon dioxide by it ceases and the level of ${ }^{14} C$ in the dead being, falls due to the decay which $C$ - 14 underoges
$ { } _6^{14} C \longrightarrow{ } _7^{14} N+\beta^{-} $
The half-life period of ${ }^{14} C$ is $5770 yr$.
The decay constant $(\lambda)$ can be calculated by using the following formula $\lambda=\frac{0.693}{t _{1 / 2}}$. The comparison of the $\beta^{-}$activity of the dead matter with that of the carbon still in circulation enables measurement of the period of the isolation of the material from the living cycle. The method however, ceases to be accurate over periods longer than 30,000 yr. The proportion of ${ }^{14} C$ to ${ }^{12} C$ in living matter is $1: 10^{12}$.
50. A nuclear explosion has taken place leading to increase in concentration of $C^{14}$ in nearby areas. $C^{14}$ concentration is $C _1$ in nearby areas and $C _2$ in areas far away. If the age of the fossil is determined to be $T _1$ and $T _2$ at the places respectively then
(a) the age of fossil will increase at the place where explosion has taken place and $T _1-T _2=\frac{1}{\lambda} \ln \frac{C _1}{C _2}$
(b) the age of fossil will decrease at the place where explosion has taken place and $T _1-T _2=\frac{1}{\lambda} \ln \frac{C _1}{C _2}$
(c) the age of fossil will be determined to be the same
(d) $\frac{T _1}{T _2}=\frac{C _1}{C _2}$
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Answer:
Correct Answer: 50. (a)
Solution:
- $\lambda T=\ln \frac{N _0}{N}$
where $N _0=$ Number of $C^{14}$ in the living matter and $N=$ Number of $C^{14}$ in fossil. Due to nuclear explosion, amount of $C^{14}$ in the near by area increases. This will increase $N _0$ because living plants are still taking $C-14$ from atmosphere, during photosynthesis, but $N$ will not change because fossil will not be doing photosynthesis.
$\Rightarrow \quad T$ (age) determined in the area where nuclear explosion has occurred will be greater than the same determined in normal area.
Also, $\lambda T _1=\ln \frac{C _1}{C} \Rightarrow \lambda T _2=\ln \frac{C _2}{C} \Rightarrow T _1-T _2=\frac{1}{\lambda}=\ln \frac{C _1}{C _2}$
$C=$ Concentration of $C-14$ in fossil.
BETA
