1.

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 upper atmosphere. ._(7)^(14)N + ._(0)^(1)n rarr ._(6)^(14)C + ._(1)n^(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 .^(14)C undergoes. ._(6)^(14)C rarr ._(7)^(14)N + beta^(-) The half-life period of .^(14)C is 5770 years. The decay constant (lamda) can be calculated by using the following formula lamda = (0.693)/(t_(1//2)). The comparison of the beta^(-) activity of the dead matter with that of the carbon still in circulation enable 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 years. The proportion of .^(14)C " to " .^(12)C in living matter is 1 : 10^(12).A nuclear explosion has taken place leading to increase in concentration of .^(14)C in nearby areas. C^(14) concentrations 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

Answer»

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))`
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))`
The age of fossil will be determined to be same
`(T_(1))/(T_(2)) = (C_(1))/(C_(2))`

SOLUTION :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))`


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