# Dating wine tritium

With the availability of accelerators like Lawrence's cyclotron or Cockroft and Walton's machine, it became possible to produce radioisotopes by bombarding elements with a beam of accelerated ions like protons, deuterons or alpha particles (helium ions).

The approach was tested in three adjacent caves in northwestern Germany which were monitored for about two years.All of the studied drip sites yielded drip water ages between 2 and 4 years with uncertainties on the order of 1 year.In contrast to traditional tritium dating, we do not refer directly to tritium concentrations in precipitation as input function, but to an infiltration-weighted annual mean of the rainwater values.Using concentration differences between this infiltration-weighted mean and the drip water, an age is calculated from the radioactive decay law, assuming piston flow.A₂/A₁ = 64 Hence, t₁ = 1 y ( 12.33 a ∙ ln(64)/ln(2) ) = 74.98 y ≈ 75 a That means the wine is 75 years old, its from 1938. Basic assumption is that the environment (soil, air, water etc) of the Amontillado vineyard contains a constant concentration of tritium - more precisely: a constant fraction of the hydrogen atoms in the environment are present as isotope tritium.

The flourishing grape-vines are in equilibrium with their environment and take the material, which they need to grow and to form fruit, from their environment.

The glass on the left has sixty-four times fewer tritium counts than the glass on the right. Tritium undergoes radioactive decay with a half-life of 12.33 years.

The glass on the left has sixty-four times fewer tritium counts than the glass on the right. What is the age of the wine in the glass on the left, and the glass on the right? What assumption(s) do you have to make in order to compare their tritium counts? Is this an accurate assumption in light of the events of the post- World War II world?

N(t) = N₀∙e^ Half-life t_½ and decay constant are related as: λ = ln(2)/t½ You can derive this relation from: (1/2)∙N₀ = N(t½) = N₀∙e^ The counter doesn't measure the number of atoms. the number of radioactive decays per unit time, which is proportional to the number of radioactive atoms: A = - d N/dt = λ∙N = λ∙N₀∙e^ Assume that both samples of wine contained the same number of atoms at the year of the production.

Then counts for each glass are given by: - left glass A₁ = λ∙N₀∙e^ - right glass A₂ = λ∙N₀∙e^ Since the exponential function decreases with time t, The gals with higher count, i.e. Therefore t₂ = 1 a When you consider the ratio of the activities the unknown N₀ cancels out: A₂/A₁ = [λ∙N₀∙e^] / [λ∙N₀∙e^] = e^/e^ = e^ ∙ e^ = e^ } t₁ = t₂ ( t_½ ∙ ln(A₂/A₁)/ln(2) ) The activity of the right glass is 64 times higher i.e.

One TU (Tritium Unit) means a tritium to hydrogen ratio of 10.