Question

For the nuclei ⁷3Li and ⁷4Be, determine the difference in binding energy per nucleon (in MeV). (Let the mass of a proton be 1.0078 u, the mass of a neutron be 1.0087 u, the mass of ⁷3Li be 7.0160 u, and the mass of ⁷4Be be 7.0169 u.)

____MeV

Answer 1

The difference in binding energy per nucleon between the nuclei ⁷³3Li and ⁷³4Be is calculated to be 0.2424 MeV/nucleon.

Step-by-step explanation:

To determine the difference in binding energy per nucleon for the nuclei 73Li and 74Be, we first need to calculate the binding energy for each nucleus using the given atomic masses of the components and the mass defect formula.

For 73Li:

• Mass of protons (Z=3) = 3 × 1.0078 u = 3.0234 u
• Mass of neutrons (N=4) = 4 × 1.0087 u = 4.0348 u
• Total mass = 3.0234 u + 4.0348 u = 7.0582 u
• Mass defect (Δm) = Total mass - Mass of 73Li = 7.0582 u - 7.0160 u = 0.0422 u
• Binding energy (BE) = Δm × 931.5 MeV/u = 0.0422 u × 931.5 MeV/u = 39.3343 MeV
• Binding energy per nucleon (BEN) = BE/A = 39.3343 MeV / 7 = 5.6192 MeV/nucleon

For 74Be:

• Mass of protons (Z=4) = 4 × 1.0078 u = 4.0312 u
• Mass of neutrons (N=3) = 3 × 1.0087 u = 3.0261 u
• Total mass = 4.0312 u + 3.0261 u = 7.0573 u
• Mass defect (Δm) = Total mass - Mass of 74Be = 7.0573 u - 7.0169 u = 0.0404 u
• Binding energy (BE) = Δm × 931.5 MeV/u = 0.0404 u × 931.5 MeV/u = 37.6374 MeV
• Binding energy per nucleon (BEN) = BE/A = 37.6374 MeV / 7 = 5.3768 MeV/nucleon

Difference in binding energy per nucleon:

ΔBEN = BENLi - BENBe = 5.6192 MeV/nucleon - 5.3768 MeV/nucleon = 0.2424 MeV/nucleon

Therefore, the difference in binding energy per nucleon between 73Li and 74Be is 0.2424 MeV/nucleon.