Contaminated groundwater at a nuclear research facility has tritium contents est
ID: 1059396 • Letter: C
Question
Contaminated groundwater at a nuclear research facility has tritium contents estimated at 10 TU (1 TU = 1 T atom in 10^18 H atoms). To make a measurement of the tritium contents by the T-^3 He grow-in method we have ~50 grams of water available. We want to make a measurement at least 10-times the mass spectrometer blank (1 times 10^-15 cm^3 STP^3 He). What is the minimum time necessary for grow-in of the requisite amount of^3 He? (Na = 6.022 times 10^23; molar volume = 22413.6 cm^3 STP; half-life of tritium = 12.26 yrs; 1 mole H_2O weights 18 g).Explanation / Answer
1 TU = 1 in 1018 H atoms
10 TU = 10 in 1018 H atoms
50g of water equal to 50x2 x 6.023 x 1023 /18 = 3.34611 x 1024 atoms of hydrogen
1018 atoms contain 10 atoms of tritium
3.34611 x 1024 atoms of hydrogen contain 10 x 3.34611 x 1024/ 1018 = 3.34611x 107 atoms of tritium
Decay of T in one day
N = Noe-t
No = 3.34611x 107 atoms
N = 3.34611x 107 e-(0.693/12.26*365) = 3.34611x 107 x 0.999845 = 3.34559x 107 atoms
No of atoms decayed = (3.34611x 107- 3.34559x 107) = 5.1815 x 103 /day
Atoms to volume conversion
1 mole (6g) of tritium contains (2x6.023x1023 atoms) and occupies 22413.6 cm3 at STP
5.1815 x 103 atoms corresponds to 9.64107x 10-17 cm3 of T2 gas
Each of T2 gas gives 2 atoms of 3He, so the volume of the He gas is 2x 9.64107x 10-17 cm3 = 1.92821 x 10-16 cm3
No of days required to produce appreciable amount of detection is 1x10-14 /1.92821x10-16 =
51.86 days = 1.7 months (51.86/30)
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