Much of the information presented in this section is based upon the Stuiver and Polach (1977) paper "Discussion: Reporting of C14 data". 1890 wood was chosen as the radiocarbon standard because it was growing prior to the fossil fuel effects of the industrial revolution.

A copy of this paper may be found in the Radiocarbon Home Page The radiocarbon age of a sample is obtained by measurement of the residual radioactivity. T (National Institute of Standards and Technology; Gaithersburg, Maryland, USA) Oxalic Acid I (C). The activity of 1890 wood is corrected for radioactive decay to 1950.

The Oxalic acid standard was made from a crop of 1955 sugar beet. The isotopic ratio of HOx I is -19.3 per mille with respect to (wrt) the PBD standard belemnite (Mann, 1983). T designation SRM 4990 C) was made from a crop of 1977 French beet molasses.

Note that the purpose of this task is algebraic in nature -- closely related tasks exist which approach similar problems from numerical or graphical stances.

The standards do not prescribe that students use or know with log identities, which form the basis for the "take the logarithm of both sides" approach.

Thereafter, the concentration (fraction) of 14C declines at a fixed exponential rate due to the radioactive decay of 14C. ) Comparing the remaining 14C fraction of a sample to that expected from atmospheric 14C allows us to estimate the age of the sample.

Raw (i.e., uncalibrated) radiocarbon ages are usually reported in radiocarbon years "Before Present" (BP), with "present" defined as CE 1950.

The task requires the student to use logarithms to solve an exponential equation in the realistic context of carbon dating, important in archaeology and geology, among other places.

Students should be guided to recognize the use of the logarithm when the exponential function has the given base of $e$, as in this problem.Now, take the logarithm of both sides to get $$ -0.693 = -5700k, $$ from which we can derive $$ k \approx 1.22 \cdot 10^.$$ So either the answer is that ridiculously big number (9.17e7) or 30,476 years, being calculated with the equation I provided and the first equation in your answer, respectively. The carbon-14 isotope would vanish from Earth's atmosphere in less than a million years were it not for the constant influx of cosmic rays interacting with molecules of nitrogen (N) into organic compounds during photosynthesis, the resulting fraction of the isotope 14C in the plant tissue will match the fraction of the isotope in the atmosphere.After plants die or are consumed by other organisms, the incorporation of all carbon isotopes, including 14C, stops.There are also trace amounts of the unstable radioisotope carbon-14 (14C) on Earth.