{"id":157,"date":"2026-04-10T02:53:13","date_gmt":"2026-04-10T02:53:13","guid":{"rendered":"https:\/\/books.gw-project.org\/age-dating-young-groundwater\/chapter\/ch073-xhtml\/"},"modified":"2026-05-12T02:01:04","modified_gmt":"2026-05-12T02:01:04","slug":"solution-exercise-6","status":"publish","type":"chapter","link":"https:\/\/books.gw-project.org\/age-dating-young-groundwater\/chapter\/solution-exercise-6\/","title":{"raw":"Solution Exercise\u00a06","rendered":"Solution Exercise\u00a06"},"content":{"raw":"
\r\n
\r\n\r\nEquation\u00a0(21)<\/span><\/a> is shown here.\r\n\r\n<\/div>\r\n
\r\n\r\n[latex]_{\\ }^{3}{He}_{trit} =[\/latex]\r\n\r\n<\/div>\r\n
\r\n\r\n[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]\r\n\r\n<\/div>\r\n
\r\n
Working from left to right through the right\u2011hand side of Equation\u00a0(21)<\/span><\/a>, values that must be calculated are [latex]R_{m}[\/latex], [latex]_{}^{3}{He}_{sol}[\/latex],<\/em> [latex]R_{sol}[\/latex],<\/em> and [latex]{Ne}_{sol}.[\/latex]<\/blockquote>\r\n<\/div>\r\n
\r\n
[latex]R_{a}[\/latex] is known (1.382 \u00d710-6<\/sup>) and, in the absence of other evidence, [latex]R_{terr}[\/latex] is assumed to be 2\u00a0\u00d7\u00a010-8<\/sup>.<\/blockquote>\r\n<\/div>\r\n
\r\n
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]\u00a0=\u00a032\u00a0percent,<\/blockquote>\r\n<\/div>\r\n
\r\n\r\nwhere:\r\n\r\n<\/div>\r\n<\/colgroup>\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
[latex]\\delta_{}^{3}{He}[\/latex]<\/th>\r\n=<\/th>\r\n[latex]\\left( \\frac{R_{m}}{R_{a}} - 1 \\right) [\/latex]100%<\/th>\r\n<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
32%<\/td>\r\n=<\/td>\r\n[latex]\\frac{R_{m}}{R_{a}}100\\% - 1(100\\%)[\/latex]<\/td>\r\n<\/tr>\r\n
0.32<\/td>\r\n=<\/td>\r\n[latex]\\frac{R_{m}}{R_{a}}1.0 - 1(1.0)[\/latex]<\/td>\r\n<\/tr>\r\n
0.32+1<\/td>\r\n=<\/td>\r\n[latex]\\frac{R_{m}}{R_{a}}[\/latex]<\/td>\r\n<\/tr>\r\n
1.32<\/td>\r\n=<\/td>\r\n[latex]\\frac{R_{m}}{R_{a}}[\/latex]<\/td>\r\n<\/tr>\r\n
[latex]R_{m}[\/latex]<\/td>\r\n=<\/td>\r\n1.32 Ra<\/sub><\/em>\u00a0= 1.32 (1.382x10-6<\/sup>)\u00a0= 1.824x10-6<\/sup><\/td>\r\n<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n
\r\n\r\n[latex]_{}^{3}{He}_{sol}[\/latex] and [latex]{Ne}_{sol}[\/latex] are determined from Equation\u00a0(4)<\/span><\/a>.\r\n\r\n<\/div>\r\n
\r\n\r\n[latex]c_{i,eq} = \\frac{\\left( P - e(T) \\right)x_{i}}{K_{w,i}(T,S)}[\/latex]\r\n\r\n<\/div>\r\n
\r\n\r\nWe need to compute P<\/em> at 500\u00a0m, e<\/em> at 17.5\u00a0\u00b0C, and [latex]K_{w,i}[\/latex] at 17.5\u00a0\u00b0C. P<\/em> is calculated as:\r\n\r\n<\/div>\r\n
\r\n\r\n[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]\r\n\r\n<\/div>\r\n
\r\n\r\ne<\/em> at 17.5\u00a0\u00b0C is calculated and converted to atmospheres as shown here.\r\n\r\n<\/div>\r\n
\r\n\r\n[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]\r\n\r\n<\/div>\r\n
\r\n\r\nTable 1<\/span><\/a> gave[latex]\\ K_{w}[\/latex] values for [latex]He[\/latex] and [latex]Ne[\/latex] at 10\u00a0\u00b0C and 20\u00a0\u00b0C but not at 17.5 \u00b0C. The empirical equations of Weiss (1971) could be used, or the values could be linearly interpolated between 10\u00a0\u00b0C and 20\u00a0\u00b0C.\r\n\r\n<\/div>\r\n
\r\n\r\nFor [latex]He,\\ with\\ K_{w,He}\\ in\\ \\frac{atm\\ g\\ }{ccSTP}[\/latex]:\r\n\r\n<\/div>\r\n
\r\n
[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]<\/blockquote>\r\n<\/div>\r\n
\r\n\r\nFor [latex]Ne[\/latex]:\r\n\r\n<\/div>\r\n
\r\n
[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]<\/blockquote>\r\n<\/div>\r\n
\r\n\r\nWe also need the mixing ratios (xi<\/sub><\/em>) of [latex]_{}^{4}{He} [\/latex]and [latex]Ne[\/latex], which are given Table\u00a04<\/span><\/a>.\r\n\r\n<\/div>\r\n
\r\n
[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\u20116<\/sup>)(0.999)[latex] = [\/latex]5.235[latex]x[\/latex]10\u20116<\/sup><\/blockquote>\r\n<\/div>\r\n
\r\n\r\n[latex]Ne[\/latex] makes up 0.001818\u00a0percent of dry atmospheric air, so xNe<\/sub>\u00a0=\u00a01.818x10-5<\/sup>.\r\n\r\n<\/div>\r\n
\r\n\r\nReturning to Equation\u00a0(4)<\/span><\/a> for [latex]He[\/latex], the concentration is calculated as shown here.\r\n\r\n<\/div>\r\n
\r\n\r\n[latex]C_{_{\\ }^{4}{He},eq} = \\frac{(0.942 - 0.0197)(5.235x10^{- 6})}{113.25\\ } = \\ 4.261x10^{- 8}\\ \\frac{ccSTP\\ }{g}[\/latex]\r\n\r\n<\/div>\r\n
\r\n\r\nAnd then we do the same for [latex]Ne[\/latex].\r\n\r\n<\/div>\r\n
\r\n\r\n[latex]C_{Ne,eq} = \\frac{(0.942 - 0.0197)(1.818x10^{- 5})}{94.25\\ } = 1.778x10^{- 7}\\ \\frac{ccSTP\\ }{g}[\/latex]\r\n\r\n<\/div>\r\n
\r\n\r\nThe last variable needed to solve Equation\u00a0(21)<\/span><\/a> is [latex]R_{sol}[\/latex]\r\n\r\n<\/div>\r\n
\r\n\r\n[latex]R_{sol} = \\ \\alpha R_{a} = 0.98\\ (1.382x10^{- 6}) = 1.354x10^{- 6}[\/latex]\r\n\r\n<\/div>\r\n
\r\n\r\nSubstituting the above values into Equation\u00a0(21)<\/span><\/a>, we calculate [latex]_{\\ }^{3}{He}_{trit}[\/latex] as follows.\r\n\r\n<\/div>\r\n
\r\n\r\n[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]\r\n\r\n<\/div>\r\n
\r\n\r\n[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]\r\n\r\n<\/div>\r\n
\r\n\r\n[latex]_{\\ }^{3}{He}_{trit} = 2.43x10^{- 14}\\ \\frac{ccSTP\\ }{g}[\/latex]\r\n\r\n<\/div>\r\n
\r\n\r\nConverting [latex]_{\\ }^{3}{He}_{trit}[\/latex] in [latex]\\frac{ccSTP\\ }{g} [\/latex]to TU\r\n\r\n<\/div>\r\n
\r\n\r\n[latex]_{\\ }^{3}{He}_{trit} = \\left( 2.43x10^{- 14} \\right)\\left( 4.021x10^{14} \\right) = 9.8\\ TU[\/latex]\r\n\r\n<\/div>\r\n
\r\n\r\nFinally, the [latex]_{}^{3}H[\/latex]\/[latex]_{}^{3}{He}[\/latex] age based on Equation\u00a0(14)<\/span><\/a> as shown here.\r\n\r\n<\/div>\r\n
\r\n\r\n[latex]Age = \\frac{\\ln\\left( 1 + \\frac{9.8}{5.6} \\right)}{\\frac{ln(2)}{12.32\\ years}} = 18\\ years[\/latex]\r\n\r\n<\/div>\r\n
\r\n\r\nReturn to Exercise\u00a06<\/span><\/a>\u2ba5\r\n\r\n<\/div>\r\n
\r\n\r\nReturn to where text linked to Exercise\u00a06<\/u><\/span><\/a>\u2ba5\r\n\r\n<\/div>\r\n<\/div>","rendered":"
\n
\n

Equation\u00a0(21)<\/span><\/a> is shown here.<\/p>\n<\/div>\n

\n

[latex]_{\\ }^{3}{He}_{trit} =[\/latex]<\/p>\n<\/div>\n

\n

[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]<\/p>\n<\/div>\n

\n

Working from left to right through the right\u2011hand side of Equation\u00a0(21)<\/span><\/a>, values that must be calculated are [latex]R_{m}[\/latex], [latex]_{}^{3}{He}_{sol}[\/latex],<\/em> [latex]R_{sol}[\/latex],<\/em> and [latex]{Ne}_{sol}.[\/latex]<\/p><\/blockquote>\n<\/div>\n

\n

[latex]R_{a}[\/latex] is known (1.382 \u00d710-6<\/sup>) and, in the absence of other evidence, [latex]R_{terr}[\/latex] is assumed to be 2\u00a0\u00d7\u00a010-8<\/sup>.<\/p><\/blockquote>\n<\/div>\n

\n

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]\u00a0=\u00a032\u00a0percent,<\/p><\/blockquote>\n<\/div>\n

\n

where:<\/p>\n<\/div>\n\n\n\n\n\n<\/colgroup>\n\n\n\n\n\n\n\n\n
[latex]\\delta_{}^{3}{He}[\/latex]<\/th>\n=<\/th>\n[latex]\\left( \\frac{R_{m}}{R_{a}} - 1 \\right)[\/latex]100%<\/th>\n<\/th>\n<\/tr>\n<\/thead>\n
32%<\/td>\n=<\/td>\n[latex]\\frac{R_{m}}{R_{a}}100\\% - 1(100\\%)[\/latex]<\/td>\n<\/tr>\n
0.32<\/td>\n=<\/td>\n[latex]\\frac{R_{m}}{R_{a}}1.0 - 1(1.0)[\/latex]<\/td>\n<\/tr>\n
0.32+1<\/td>\n=<\/td>\n[latex]\\frac{R_{m}}{R_{a}}[\/latex]<\/td>\n<\/tr>\n
1.32<\/td>\n=<\/td>\n[latex]\\frac{R_{m}}{R_{a}}[\/latex]<\/td>\n<\/tr>\n
[latex]R_{m}[\/latex]<\/td>\n=<\/td>\n1.32 Ra<\/sub><\/em>\u00a0= 1.32 (1.382×10-6<\/sup>)\u00a0= 1.824×10-6<\/sup><\/td>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n
\n

[latex]_{}^{3}{He}_{sol}[\/latex] and [latex]{Ne}_{sol}[\/latex] are determined from Equation\u00a0(4)<\/span><\/a>.<\/p>\n<\/div>\n

\n

[latex]c_{i,eq} = \\frac{\\left( P - e(T) \\right)x_{i}}{K_{w,i}(T,S)}[\/latex]<\/p>\n<\/div>\n

\n

We need to compute P<\/em> at 500\u00a0m, e<\/em> at 17.5\u00a0\u00b0C, and [latex]K_{w,i}[\/latex] at 17.5\u00a0\u00b0C. P<\/em> is calculated as:<\/p>\n<\/div>\n

\n

[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]<\/p>\n<\/div>\n

\n

e<\/em> at 17.5\u00a0\u00b0C is calculated and converted to atmospheres as shown here.<\/p>\n<\/div>\n

\n

[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]<\/p>\n<\/div>\n

\n

Table 1<\/span><\/a> gave[latex]\\ K_{w}[\/latex] values for [latex]He[\/latex] and [latex]Ne[\/latex] at 10\u00a0\u00b0C and 20\u00a0\u00b0C but not at 17.5 \u00b0C. The empirical equations of Weiss (1971) could be used, or the values could be linearly interpolated between 10\u00a0\u00b0C and 20\u00a0\u00b0C.<\/p>\n<\/div>\n

\n

For [latex]He,\\ with\\ K_{w,He}\\ in\\ \\frac{atm\\ g\\ }{ccSTP}[\/latex]:<\/p>\n<\/div>\n

\n

[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]<\/p><\/blockquote>\n<\/div>\n

\n

For [latex]Ne[\/latex]:<\/p>\n<\/div>\n

\n

[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]<\/p><\/blockquote>\n<\/div>\n

\n

We also need the mixing ratios (xi<\/sub><\/em>) of [latex]_{}^{4}{He}[\/latex]and [latex]Ne[\/latex], which are given Table\u00a04<\/span><\/a>.<\/p>\n<\/div>\n

\n

[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\u20116<\/sup>)(0.999)[latex]=[\/latex]5.235[latex]x[\/latex]10\u20116<\/sup><\/p><\/blockquote>\n<\/div>\n

\n

[latex]Ne[\/latex] makes up 0.001818\u00a0percent of dry atmospheric air, so xNe<\/sub>\u00a0=\u00a01.818×10-5<\/sup>.<\/p>\n<\/div>\n

\n

Returning to Equation\u00a0(4)<\/span><\/a> for [latex]He[\/latex], the concentration is calculated as shown here.<\/p>\n<\/div>\n

\n

[latex]C_{_{\\ }^{4}{He},eq} = \\frac{(0.942 - 0.0197)(5.235x10^{- 6})}{113.25\\ } = \\ 4.261x10^{- 8}\\ \\frac{ccSTP\\ }{g}[\/latex]<\/p>\n<\/div>\n

\n

And then we do the same for [latex]Ne[\/latex].<\/p>\n<\/div>\n

\n

[latex]C_{Ne,eq} = \\frac{(0.942 - 0.0197)(1.818x10^{- 5})}{94.25\\ } = 1.778x10^{- 7}\\ \\frac{ccSTP\\ }{g}[\/latex]<\/p>\n<\/div>\n

\n

The last variable needed to solve Equation\u00a0(21)<\/span><\/a> is [latex]R_{sol}[\/latex]<\/p>\n<\/div>\n

\n

[latex]R_{sol} = \\ \\alpha R_{a} = 0.98\\ (1.382x10^{- 6}) = 1.354x10^{- 6}[\/latex]<\/p>\n<\/div>\n

\n

Substituting the above values into Equation\u00a0(21)<\/span><\/a>, we calculate [latex]_{\\ }^{3}{He}_{trit}[\/latex] as follows.<\/p>\n<\/div>\n

\n

[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]<\/p>\n<\/div>\n

\n

[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]<\/p>\n<\/div>\n

\n

[latex]_{\\ }^{3}{He}_{trit} = 2.43x10^{- 14}\\ \\frac{ccSTP\\ }{g}[\/latex]<\/p>\n<\/div>\n

\n

Converting [latex]_{\\ }^{3}{He}_{trit}[\/latex] in [latex]\\frac{ccSTP\\ }{g}[\/latex]to TU<\/p>\n<\/div>\n

\n

[latex]_{\\ }^{3}{He}_{trit} = \\left( 2.43x10^{- 14} \\right)\\left( 4.021x10^{14} \\right) = 9.8\\ TU[\/latex]<\/p>\n<\/div>\n

\n

Finally, the [latex]_{}^{3}H[\/latex]\/[latex]_{}^{3}{He}[\/latex] age based on Equation\u00a0(14)<\/span><\/a> as shown here.<\/p>\n<\/div>\n

\n

[latex]Age = \\frac{\\ln\\left( 1 + \\frac{9.8}{5.6} \\right)}{\\frac{ln(2)}{12.32\\ years}} = 18\\ years[\/latex]<\/p>\n<\/div>\n

\n

Return to Exercise\u00a06<\/span><\/a>\u2ba5<\/p>\n<\/div>\n

\n

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