Answers · Sharp, Principles of Stable Isotope Geochemistry, 1st ed. (2007) · ← Problems
From equation (4.16), with Δ(N2 gas–nitrate) = −35‰, α = 0.965:
The N2 gas will be 35‰ less than the dissolved nitrate.
As with carbon, availability affects the fractionation during the conversion from one form to another. If there is very little ammonium available, there can be very little discrimination. The algae takes “whatever it can get.” If there is excess ammonium, then the algae can discriminate, in which case they preferentially take the light isotopologues. In general, more food available implies more discrimination.
Use equation (4.16) to answer the question. We have δ15Nnitrate = (δ15Nnitrate,initial + 1000)·F(α−1) − 1000, where Δ(N2 gas–nitrate) is the fractionation between product and reactant (nitrate). Start with a value of 5.25 (the highest-concentration sample) and F = conc/39.7 (where 39.7 is the highest concentration). Then test different values of α to see what fits the data best. The simple Rayleigh distillation equation can also be linearized, so that a simple linear regression can be applied to the data. See Sharp et al. (2003) for details.
[Worked grid (constructed), matching the worksheet in the problem set. The linearized regression gives slope α − 1 = −0.0349, i.e. α = 0.965 and Δ15N(N2 gas–nitrate) ≈ −35‰, in agreement with the printed route of testing trial α values. Click a cell to see its formula; the plot opens on the linearized route with the fit and the model curve for the fitted α.]
Keep buying apples! A δ15N value of +16‰ is consistent with an organic fertilizer nitrogen source. See Fig. 9.7.
Here is a classic problem of steady state. The assimilation of nitrogen in the ocean has no fractionation, but the N2 leaving the ocean has a positive discrimination of +20‰. That is, the δ15N of the N2 leaving the ocean is 20‰ heavier than the δ15N of the ocean itself. In order to have a steady-state balance, the δ15N value entering the ocean must equal the δ15N value leaving the ocean; otherwise the ocean value would change with time. If the fractionation is +20‰, then the ocean must have an average δ15N value of −20‰ for the N2 leaving to have the same value as what enters. reconstructed [The printed final value is lost in the extraction; −20‰ follows from the printed steady-state argument.]
An Eskimo community should have very high δ15N values and intermediate δ13C values. The diet comes from a very high trophic level (seal, etc.), making the δ15N value high, whereas carbon is of marine origin, intermediate between C3 and C4 plants.
Big increase. The minor/major isotope ratio of carbon in CO is 1.11/98.89, vs. 0.37/99.63 for nitrogen in N2. [The nitrogen digits are garbled in the extraction (printed text reads “99.759”); check vs print.] Calculate the apparent increase for a 5% contamination by CO of a typical “air” N2 value.
Transcribed from Z. Sharp, Principles of Stable Isotope Geochemistry, 1st ed. (Pearson Prentice Hall, 2007), end-of-chapter problem set (published source). Reference-scale names are written VSMOW, VPDB and VCDT where the original used SMOW, PDB and CDT, with the original numerical constants unchanged; historical usages (e.g. Craig's definition of SMOW) are kept as printed.
The source PDF's text layer scrambles mathematics, so equations and some values were reconstructed during transcription. Bracketed notes and anything tagged reconstructed or check vs print are editorial (constructed), not part of the original, and the warn-tagged items should be verified against the print copy.