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 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 toincrease 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

Answer»

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

Solution :`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))`


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