A \(26.00-g\) sample of water containing tritium, \({ }_{1}^{3} \mathrm{H},\) emits \(1.50 \times 10^{3}\) beta particles per second. Tritium is a weak beta emitter with a half-life of \(12.3 \mathrm{yr}\). What fraction of all the hydrogen in the water sample is tritium?

We know that the decay rate of a radioactive isotope is proportional to the number of atoms of the isotope. Therefore, we can use the following formula to find out the decay constant, λ, for tritium: \[\lambda = \frac{ln(2)}{t_{\frac{1}{2}}}\] Where \(t_{\frac{1}{2}}\) is the half-life of the isotope.

We are given the half-life of tritium, \({t_{\frac{1}{2}} = 12.3}\) years. Now, use the above formula to find the decay constant: \[\lambda = \frac{ln(2)}{12.3}\] \[\lambda \approx 0.0563 \mathrm{yr}^{-1}\]

We are given that the tritium in the sample emits 1.50 x \(10^3\) beta particles per second. This is its activity (number of decays per unit time). To account for the difference in units (seconds and years), we'll convert the activity to decays per year using the following conversion: \[A = 1.50 \times 10^3 \frac{\mathrm{decays}}{\mathrm{s}} \times \frac{3.1536 \times 10^7 \mathrm{s}}{\mathrm{yr}}\] \[A \approx 4.73 \times 10^{10} \mathrm{decays/yr}\]

We can calculate the number of tritium atoms, N, using the following formula, which relates activity, decay constant, and number of atoms: \[A = \lambda N\] We have already calculated A and λ, so we can find the number of tritium atoms: \[N = \frac{A}{\lambda}\] \[N \approx 8.4 \times 10^{11} \mathrm{atoms}\]

We are given that the mass of the water sample is 26.00 g. Water has a molecular formula \(H_2O\), which means there are two hydrogen atoms for each water molecule. We can find the total number of hydrogen atoms in the sample using the following steps: 1. Calculate the number of moles of water: \(\frac{26.00 g}{18.015 g/mol} = 1.44 mol\), where 18.015 g/mol is the molar mass of water. 2. Multiply the number of moles of water by Avogadro's number to find the number of water molecules: \(1.44 mol \times 6.022 \times 10^{23} mol^{-1} = 8.67 \times 10^{23}\) water molecules. 3. Since there are two hydrogen atoms in each water molecule, multiply the number of water molecules by two to get the total number of hydrogen atoms: \(8.67 \times 10^{23} \times 2 = 1.73 \times 10^{24}\) hydrogen atoms.

Now we can calculate the fraction of the tritium in the sample by dividing the number of tritium atoms by the total number of hydrogen atoms: \[\mathrm{Fraction\: of\: tritium} = \frac{N}{\mathrm{Total\: hydrogen\: atoms}}\] \[\mathrm{Fraction\: of\: tritium} = \frac{8.4 \times 10^{11}}{1.73 \times 10^{24}}\] \[\mathrm{Fraction\: of\: tritium} \approx 4.85 \times 10^{-13}\] Thus, the fraction of tritium in the water sample is approximately \(4.85 \times 10^{-13}\).