Carbon 14 is used to determine the age of organic material. The procerdure is based on the formation of `.^(14)C` by neutron capture in the upper atmosphere.
`._(7)^(14)N + ._(0)^(1)n rarr ._(6)^(14)C + ._(1)n^(1)`
`.^(14)C` is abosorbed by living organisms during phostosythesis. The `.^(14)C` content is constant in living organisms 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 `.^(14)C` undergoes.
`._(6)^(14)C rarr ._(7)^(14)C + beta^(-)`
The half-life period of `.^(14)C` is 5770 years. The decay constant `(lambda)` can be calculated by using the following formula `lambda = (0.693)/(t_(1//2))`
The comparison fo the `beta^(-)` activity fo the dead matter with that of the carbon still in circulation enables measurement of the period of the isolation the materail form the living cycle. The method however, ceases to be accurate ever periods longer than `30,000` years. The proportaion of `.^(14)C` to `.^(12)C` living matter is `1:10^(12)`.
A nulcear explosion has taken place leading to increases in conventration of `^(14)C` in nearly areas. `^(14)C` concentration is `C_(1)` in nearby areas and `C_(2)` in areas far away. If the age of the fossil is detemined to be `T_(1)` and `T_(2)` at the places respectively, then:

Options

A. The age of the fossil will increase at the place where explosion has taken and `T_(1) - T_(2) = (1)/(lamda "ln" (C_(1))/(C_(2))`
B. The age of the fossil will decrease at the place where explosion has taken place and `T_(1) - T_(2) = (1)/(lamda) "ln" (C_(1))/(C_(2))`
C. The age of fossil will be determined to be same
D. `(T_(1))/(T_(2)) = (C_(1))/(C_(2))`

Answer

A

Explanation

All radioactive decays are the examples of first order kinetics
So decay constant `lamda = (1)/(T_(1) - T_(2)) "ln" (C_(1))/(C_(2))`
`C_(1)` is the concentration at `T_(1)` times
`C_(2)` is the concentration at `T_(2)` time
So `T_(1) - T_(2) = (1)/(lamda) ln (C_(1))/(C_(2))`