The binding energy of helium 4 He is 2 [atomic mass of proton = 1.00783U, Atomic mass of neutron = 1.00867U]

Answer Details

The binding energy of helium refers to the amount of energy required to separate the two protons and two neutrons in the helium nucleus. The table provided shows the atomic masses of helium with two different numbers of neutrons. By subtracting the atomic mass of the helium nucleus with two neutrons from the one with four neutrons, we can find the mass defect, which is the mass that is converted to energy in the process of binding the nucleus. Using Einstein's famous equation E=mc², we can calculate the binding energy from the mass defect. The mass defect can be calculated as: mass defect = (atomic mass of helium with 2 neutrons) + 2 × (atomic mass of a proton) + 2 × (atomic mass of a neutron) - atomic mass of helium with 4 neutrons Substituting the given values, we get: mass defect = 2.014101u + 2 × 1.00783u + 2 × 1.00867u - 4.002603u mass defect = 0.030379u Using E=mc², we can find the binding energy, where c is the speed of light in a vacuum (299,792,458 m/s): binding energy = mass defect × c² binding energy = 0.030379u × (299,792,458 m/s)² binding energy ≈ 2.73 × 10⁻¹² J Therefore, the answer is 0.033U.