Physics Problem: Binding Energy of Lithium-7
📘 Question:
(5 pts)
Calculate the binding energy of the nucleus _3^7Li.
The atomic mass of the neutral lithium-7 atom is 7.016003 u.
Given:
- Mass of proton = 1.007276 u
- Mass of neutron = 1.008665 u
- 1 atomic mass unit (u) = 931.494 MeV
Options:
- a) 37.7 MeV
- b) 39.2 MeV
- c) 40.7 MeV
- d) 48.2 MeV
- e) None of the above
The binding energy is the energy required to break a nucleus into its individual protons and neutrons. We use the concept of mass defect and Einstein’s mass-energy equivalence formula.
Step 1: Determine Nucleus Composition
Lithium-7 contains:
- 3 protons
- 4 neutrons
- Total mass = 7.016003 u (includes electrons)
Step 2: Calculate Mass of Individual Nucleons
Mass of 3 protons: 3 × 1.007276 u = 3.021828 u
Mass of 4 neutrons: 4 × 1.008665 u = 4.034660 u
Total separate mass: 3.021828 + 4.034660 = 7.056488 u
Step 3: Subtract Electron Mass to Find Nuclear Mass
Electrons contribute: 3 × 0.000549 u = 0.001647 u
Nuclear mass = 7.016003 u − 0.001647 u = 7.014356 u
Step 4: Calculate Mass Defect
Mass defect = Mass of nucleons − Actual nuclear mass
= 7.056488 u − 7.014356 u = 0.042132 u
Step 5: Convert Mass Defect to Binding Energy
Using E = Δm × 931.494 MeV/u
Binding Energy = 0.042132 u × 931.494 MeV/u ≈ 39.25 MeV
Rounded to 1 decimal place: 39.2 MeV
✅ Final Answer:
39.2 MeV
Correct option: (b)