Chemical Kinetics - Result Question 50
####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. $T$
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.
