Exercise 6

A groundwater sample was analyzed for [latex]He,\ Ne[/latex], and [latex]_{}^{3}H[/latex] with the following results:

[latex]_{}^{4}{He} = 5.29 \times[/latex] 10^–8 ccSTP/g

[latex]\ \ Ne = 2.14 \times[/latex] 10^–7 ccSTP/g

[latex]\delta_{}^{3}{He} = 32[/latex] %

[latex]_{}^{3}{He} = 5.6[/latex] TU

The sample is thought to have recharged at an elevation of 500 m and the mean annual temperature at the site is 17.5 °C.

Calculate the [latex]\frac{_{}^{3}H}{_{}^{3}{He}}[/latex] tracer age assuming no fractionation of the excess air component. Assume that the [latex]He/Ne[/latex] ratio of excess air is 0.2882.

A useful unit conversion: 1 ccSTP/g = 4.021 × 10¹⁴ TU.

Solution to Exercise 6⮧

Return to where text linked to Exercise 6⮥