The binding energy per nucleon for magnesium- 27 is $1.326 \times 10^{-12} \mathrm{J} /\( nucleon. Calculate the atomic mass of \)^{27} \mathrm{Mg}$ .

First, we need to determine the total binding energy of magnesium-27. Since there are 27 nucleons in Mg-27, we can multiply the binding energy per nucleon by the number of nucleons: Total binding energy = Binding energy per nucleon × Number of nucleons Total binding energy = \(1.326 * 10^{-12} J/nucleon * 27\) Total binding energy = \(3.5802 * 10^{-11} J\)

Next, we'll calculate the mass defect (i.e., the difference between the sum of nucleon masses and the atomic mass of magnesium-27) using the mass-energy equivalence formula: Mass defect = Total binding energy / \(c^2\) Mass defect = \(\frac{3.5802 * 10^{-11} J}{(3 * 10^8 m / s)^2}\) Mass defect = \(3.989 * 10^{-28} kg\)

Now, we'll determine the atomic mass of magnesium-27 by adding the mass defect to the sum of the individual nucleon masses. Since magnesium has 12 protons and 15 neutrons, we need to account for their respective masses: - Mass of proton = \(1.6726 * 10^{-27} kg\) - Mass of neutron = \(1.6749 * 10^{-27} kg\) Sum of nucleon masses = (12 * Mass of proton) + (15 * Mass of neutron) Sum of nucleon masses = (\(12 * 1.6726 * 10^{-27} kg\)) + (\(15 * 1.6749 * 10^{-27} kg\)) Sum of nucleon masses = \(4.4733 * 10^{-26} kg\) Now we can calculate the atomic mass of magnesium-27: Atomic mass of Mg-27 = Sum of nucleon masses - Mass defect Atomic mass of Mg-27 = \(4.4733 * 10^{-26} kg - 3.989 * 10^{-28} kg\) Atomic mass of Mg-27 = \(4.4335 * 10^{-26} kg\) Therefore, the atomic mass of magnesium-27 is approximately \(4.4335 * 10^{-26} kg\).