Solution Exercise 6
Equation (21) is shown here.
[latex]_{\ }^{3}{He}_{trit} =[/latex]
[latex]_{\ }^{4}{He}_{m}\left( R_{m} - R_{terr} \right) -_{\ }^{4}{He}_{sol}\left( R_{sol} - R_{terr} \right) - \left( \frac{_{\ }^{4}{He}}{\ \ Ne} \right)_{EA}\left( {Ne}_{m} - {Ne}_{sol} \right)\left( R_{a} - R_{terr} \right)[/latex]
Working from left to right through the right‑hand side of Equation (21), values that must be calculated are [latex]R_{m}[/latex], [latex]_{}^{3}{He}_{sol}[/latex], [latex]R_{sol}[/latex], and [latex]{Ne}_{sol}.[/latex]
[latex]R_{a}[/latex] is known (1.382 ×10-6) and, in the absence of other evidence, [latex]R_{terr}[/latex] is assumed to be 2 × 10-8.
Measured concentrations of [latex]_{}^{4}{He}[/latex] and [latex]Ne[/latex], and the assumed [latex]_{}^{4}{He}[/latex]/[latex]Ne[/latex] ratio, are given in the problem statement. [latex]R_{m}[/latex] can be determined from [latex]\delta_{}^{3}{He}[/latex] = 32 percent,
where:
| [latex]\delta_{}^{3}{He}[/latex] | = | [latex]\left( \frac{R_{m}}{R_{a}} - 1 \right)[/latex]100% | |
|---|---|---|---|
| 32% | = | [latex]\frac{R_{m}}{R_{a}}100\% - 1(100\%)[/latex] | |
| 0.32 | = | [latex]\frac{R_{m}}{R_{a}}1.0 - 1(1.0)[/latex] | |
| 0.32+1 | = | [latex]\frac{R_{m}}{R_{a}}[/latex] | |
| 1.32 | = | [latex]\frac{R_{m}}{R_{a}}[/latex] | |
| [latex]R_{m}[/latex] | = | 1.32 Ra = 1.32 (1.382×10-6) = 1.824×10-6 | |
[latex]_{}^{3}{He}_{sol}[/latex] and [latex]{Ne}_{sol}[/latex] are determined from Equation (4).
[latex]c_{i,eq} = \frac{\left( P - e(T) \right)x_{i}}{K_{w,i}(T,S)}[/latex]
We need to compute P at 500 m, e at 17.5 °C, and [latex]K_{w,i}[/latex] at 17.5 °C. P is calculated as:
[latex]P = P_{0}\exp\left( - \frac{Z}{Z_{s}} \right) = 1\ atm\ exp\left( - \frac{500\ m}{8300\ m} \right)\ = \ 0.942\ atm[/latex]
e at 17.5 °C is calculated and converted to atmospheres as shown here.
[latex]e = 10^{A - \frac{B}{C + T}} = 10^{\left( 8.07131 - \ \frac{1730.63}{233.426 + 17.5} \right)} = 14.94\ mmHg\ \frac{1atm}{760\ mmHg} = 0.0197\ atm[/latex]
Table 1 gave[latex]\ K_{w}[/latex] values for [latex]He[/latex] and [latex]Ne[/latex] at 10 °C and 20 °C but not at 17.5 °C. The empirical equations of Weiss (1971) could be used, or the values could be linearly interpolated between 10 °C and 20 °C.
For [latex]He,\ with\ K_{w,He}\ in\ \frac{atm\ g\ }{ccSTP}[/latex]:
[latex]K_{w,He}\ at\ 17.5\ {^\circ}C\ = \ K_{w,He}\ at\ 10\ {^\circ}C\ + \frac{(17.5{^\circ}C - 10{^\circ}C)}{(20{^\circ}C - 10{^\circ}C)}\ \left( K_{w,He}\ at\ 20\ {^\circ}C\ - \ K_{w,He}\ at\ 10\ {^\circ}C \right) = \ 111\ + \frac{(17.5\ {^\circ}C - 10\ {^\circ}C)}{(20\ {^\circ}C - 10\ {^\circ}C)}(114\ - \ 111) = \ 113.25\ \frac{atm\ g\ }{ccSTP}[/latex]
For [latex]Ne[/latex]:
[latex]K_{w,Ne}\ at\ 17.5\ {^\circ}C\ = \ K_{w,Ne}\ at\ 10\ {^\circ}C\ + \frac{(17.5\ {^\circ}C - 10\ {^\circ}C)}{(20\ {^\circ}C - 10\ {^\circ}C)}\ \left( K_{w,Ne}\ at\ 20\ {^\circ}C\ - \ K_{w,Ne}\ at\ 10\ {^\circ}C \right) = \ 89.0\ + \frac{(17.5\ {^\circ}C - 10\ {^\circ}C)}{(20\ {^\circ}C - 10\ {^\circ}C)}(96.0\ - \ 89.0) = \ 94.25\ \frac{atm\ g\ }{ccSTP}[/latex]
We also need the mixing ratios (xi) of [latex]_{}^{4}{He}[/latex]and [latex]Ne[/latex], which are given Table 4.
[latex]_{}^{4}{He}[/latex] makes up 99.9 percent of atmospheric [latex]He[/latex], and [latex]He[/latex]makes up 0.000524 percent of dry atmospheric air, so [latex]x_{He}\ =[/latex](5.24[latex]x[/latex]10‑6)(0.999)[latex]=[/latex]5.235[latex]x[/latex]10‑6
[latex]Ne[/latex] makes up 0.001818 percent of dry atmospheric air, so xNe = 1.818×10-5.
Returning to Equation (4) for [latex]He[/latex], the concentration is calculated as shown here.
[latex]C_{_{\ }^{4}{He},eq} = \frac{(0.942 - 0.0197)(5.235x10^{- 6})}{113.25\ } = \ 4.261x10^{- 8}\ \frac{ccSTP\ }{g}[/latex]
And then we do the same for [latex]Ne[/latex].
[latex]C_{Ne,eq} = \frac{(0.942 - 0.0197)(1.818x10^{- 5})}{94.25\ } = 1.778x10^{- 7}\ \frac{ccSTP\ }{g}[/latex]
The last variable needed to solve Equation (21) is [latex]R_{sol}[/latex]
[latex]R_{sol} = \ \alpha R_{a} = 0.98\ (1.382x10^{- 6}) = 1.354x10^{- 6}[/latex]
Substituting the above values into Equation (21), we calculate [latex]_{\ }^{3}{He}_{trit}[/latex] as follows.
[latex]_{\ }^{3}{He}_{trit} =_{\ }^{4}{He}_{m}\left( 1.824x10^{- 6} - 2x10^{- 8} \right) -_{\ }^{4}{He}_{sol}\left( 1.354x10^{- 6} - 2x10^{- 8} \right) - (0.2882)\left( {Ne}_{m} - {Ne}_{sol} \right)\left( 1.382x10^{- 6} - 2x10^{- 8} \right)[/latex]
[latex]_{\ }^{3}{He}_{trit} = 5.29x10^{- 8}\left( 1.824x10^{- 6} - 2x10^{- 8} \right) - \left( 4.261x10^{- 8} \right)\left( 1.354x10^{- 6} - 2x10^{- 8} \right) - (0.2882)\left( 2.14x10^{- 7} - 1.778x10^{- 7} \right)\left( 1.382x10^{- 6} - 2x10^{- 8} \right)[/latex]
[latex]_{\ }^{3}{He}_{trit} = 2.43x10^{- 14}\ \frac{ccSTP\ }{g}[/latex]
Converting [latex]_{\ }^{3}{He}_{trit}[/latex] in [latex]\frac{ccSTP\ }{g}[/latex]to TU
[latex]_{\ }^{3}{He}_{trit} = \left( 2.43x10^{- 14} \right)\left( 4.021x10^{14} \right) = 9.8\ TU[/latex]
Finally, the [latex]_{}^{3}H[/latex]/[latex]_{}^{3}{He}[/latex] age based on Equation (14) as shown here.
[latex]Age = \frac{\ln\left( 1 + \frac{9.8}{5.6} \right)}{\frac{ln(2)}{12.32\ years}} = 18\ years[/latex]