The natural abundance of \(^{3} \mathrm{He}\) is 0.000137\(\%\) . (a) How many protons, neutrons, and electrons are in an atom of \(^{3} \mathrm{He?}\) (b) Based on the sum of the masses of their subatomic particles, which is expected to be more massive, an atom of \(^{3} \mathrm{He}\) or an atom of \(^{3} \mathrm{H}(\) which is also called tritium)? (c) Based on your answer to part (b), what would need to be the precision of a mass spectrometer that is able to differentiate between peaks that are due to \(^{3} \mathrm{He}^{+}\) and \(^{3} \mathrm{H}^{+}\) ?

We can use the isotopic notation to determine the number of protons, neutrons, and electrons in an atom of \(^{3}\mathrm{He}\). The isotope notation for He-3 is written as \(_{2}^{3}\mathrm{He}\), where 2 is the atomic number (Z) representing the number of protons, and 3 is the mass number (A) representing the total number of protons and neutrons. From the notation: Number of protons (Z) = 2 Number of neutrons (N) = A - Z = 3 - 2 = 1 Since the atom is neutral, the number of electrons equals the number of protons: Number of electrons = 2 Thus, there are 2 protons, 1 neutron, and 2 electrons in an atom of \(^{3}\mathrm{He}\).

Now, let's compare the masses of \(^{3}\mathrm{He}\) and \(^{3}\mathrm{H}\) based on the sum of their subatomic particles. Given the masses of proton, neutron, and electron as: Mass of a proton (m_p) = \(1.0073 \ \mathrm{amu}\), Mass of a neutron (m_n) = \(1.0087 \ \mathrm{amu}\), and Mass of an electron (m_e) = \(5.486 \times 10^{-4} \ \mathrm{amu}\). We can then calculate the approximate masses for \(^{3}\mathrm{He}\) and \(^{3}\mathrm{H}\). For \(^{3}\mathrm{He}\): mass = (number of protons)×(mass of proton) + (number of neutrons)×(mass of neutron) + (number of electrons)×(mass of electron) mass = (2×1.0073 + 1×1.0087 + 2×5.486 \times 10^{-4})\ \mathrm{amu} mass ≈ 3.0157\ \mathrm{amu} For \(^{3}\mathrm{H}\), let's denote the number of neutrons by N': Number of neutrons (N') = mass number of \(^{3}\mathrm{H}\) - Z = 3 - 1 = 2 Mass of \(^{3}\mathrm{H}\): mass = (number of protons)×(mass of proton) + (number of neutrons)×(mass of neutron) + (number of electrons)×(mass of electron) mass = (1×1.0073 + 2×1.0087 + 1×5.486 \times 10^{-4})\ \mathrm{amu} mass ≈ 3.0160\ \mathrm{amu} Thus, the atom of \(^{3}\mathrm{H}\) is expected to be more massive than the atom of \(^{3}\mathrm{He}\).

To differentiate between the peaks due to \(^{3}\mathrm{He}^{+}\) and \(^{3}\mathrm{H}^{+}\), the mass spectrometer should be able to resolve masses with a precision equal to or better than the mass difference between these ions. Considering that their difference comes from only one additional neutron in the case of \(^{3}\mathrm{H}^{+}\), we can calculate the required precision as Required precision = \(\frac{\text{mass difference}}{\text{average mass}}\) Since we are concerned only with the ions, we can ignore the mass of the electrons: mass difference = \(|\text{mass}(^{3}\mathrm{H}^{+}) - \text{mass}(^{3}\mathrm{He}^{+})|\) where mass(\(^{3}\mathrm{H}^{+}\)) includes the mass of 1 proton and 2 neutrons, and mass(\(^{3}\mathrm{He}^{+}\)) includes the mass of 2 protons and 1 neutron. mass difference = \(|(1×1.0073 + 2×1.0087) - (2×1.0073 + 1×1.0087)|\ \mathrm{amu}\) mass difference = \(0.0013\ \mathrm{amu}\) To find the average mass, we can average the mass of \(^{3}\mathrm{H}^{+}\) and \(^{3}\mathrm{He}^{+}\): average mass = \(\frac{\text{mass}(^{3}\mathrm{H}^{+}) + \text{mass}(^{3}\mathrm{He}^{+})}{2}\) average mass = \(\frac{(1×1.0073 + 2×1.0087) + (2×1.0073 + 1×1.0087)}{2}\ \mathrm{amu}\) average mass = 3.0035\ \mathrm{amu} Now, we can calculate the required precision: required precision = \(\frac{0.0013\ \mathrm{amu}}{3.0035\ \mathrm{amu}}\) required precision ≈ 0.000433 Hence, to differentiate between peaks due to \(^{3}\mathrm{He}^{+}\) and \(^{3}\mathrm{H}^{+}\), a mass spectrometer should have a precision of approximately 0.000433 or better.