Carbon - 14 is used to determine the age of organic material . The procedure is based on the formation of `""^(14) C` by neutron capture in the atmosphere . `T_(1) - T_(2) = (1)/(lambda) "log" (C_(1))/(C_(2))`
`""^(14) C` is absorbed by living organisms during photosynthesis . The `""^(14) C` content is content is constant in living organism once the plant or animal dies , the uptake of carbon dioxide by it increases and the level of `""^(14) C` in the dead being falls due to the decay which `C^(14)` undergoes .
`""_(6)^(14) C to ""_(7)^(14) N + beta^(-) (""_(-1)^(0)e)` . The half- life period of `""^(14) C` is 5770 years . The decay constant `(lambda)` can be calculated using the formula `lambda = (0.693)/(t_(1//2))`.
The comparison of the `b^(-)` 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 period longer than 30,000 years . The proportion of `""^(14) C` to `""^(12) C` in living matter is `1 : 10^(2)`
A nuclear explosion has taken place leading to increase in concentration of `C^(14)` is `C_(3)` 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

Options

A. the age of the fossil will increase at the place where explosion has taken place and `T_(1) - T_(2) = (1)/(lambda) "log" (C_(1))/(C_(2))`
B. the age of the fossil will decreases at the place where explosion has taken place and `T_(1) - T_(2) = (1)/(lambda) "log" (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

`lambda T_(1) = 1_n ((c_(1))/(c_(2))) …… (1) lambda T_(2) = 1_(n) ((c_(2))/(c_(1)))….. (2)`
`lambda (T_(1) - T_(2)) = 1_(n) ((c_1)/(c_1)) - 1_n ((c_2)/(c_1)) , lambda (T_(1) -T_(2)) = 1_(n) ((c_(1))/(c_(2)) xx (c_(1))/(c_(2))) , (T_(1) - T_(2)) = (1)/(lambda) xx 1_n ((c_1)/(c_2))`