Question

The radioactive decay of carbon-14 is first order and the half-life is 5800 years. While a plant or animal is living, it has a constant proportion of carbon-14 (relative to carbon-12) in its composition. When the organism dies, the proportion of carbon-14 decreases as a result of radioactive decay and the age of the organism can be determined if the proportion of carbon-14 in its remains is measured. If the proportion of carbon-14 in an ancient piece of wood is found to be one quarter that in living trees, how old is the sample?

Answer 1

if it has a half life of 5800, it means that 100% of the carbon-14 will be 50%. after 5800 years


so if it goes through 2 half lifes it will be 25% or 1/4 of the living subtance. so 5800*2 =11,600

Answer 2

Since this is a radioactive disintegration and it is stated, the reaction is first order. For a first order reaction, the half life (t subscript 1/2)should be equal to ln(2) / k. Since the half life is given, k can be calculated. :) Okay, then for the amount of carbon-14 remaining, the equation would be C = Co exp (-kt) where k is determined before. using the relationship C = 1/4 of Co C now is one-quarter that of the one in living trees, t can be calculated, which is now 11,600 years. This would actually make sense since the half life is 5800 and two half-lives have passed already, thus. two 5800 years (which is 11600 :)))