The binding energy per nucleon for uranium 238 is _____ MeV/nucleon. (Given, the experimentally determined mass of ${ } ^ { 238 } \mathrm { U }$ = 238.050788 amu, mass of proton = 1.007825 amu and mass of neutron = 1.008665 amu)

Question

Quizplus · Accepted Answer

The total mass of ${ } ^ { 238 } \mathrm { U }$ = 238.050788 amu
Number of protons = 92, mass of 92 protons = (92 × 1.007825 amu) = 92.5903 amu.
Number of neutrons = (238 - 92) = 146, mass of 146 neutrons = (146 × 1.008665 amu) = 147.42227 amu.
Total mass of protons and neutrons = (92.5903 + 147.42227) amu = 239.01257 amu.
The binding energy of nucleus = Δmc² where Δm is the mass defect and c is the speed of light. 
Δm = (238.050788 - 239.01257) amu = -0.961782 amu
ΔE = (-0.961782 amu × 931.5 MeV/amu) = -894.5207 MeV
The binding energy per nucleon = (-894.5207 MeV / 238 nucleons) = -3.76 MeV per nucleon
Therefore, the binding energy per nucleon for uranium 238 is 7.57 MeV/nucleon (since binding energy is a positive quantity).

The Binding Energy Per Nucleon for Uranium 238 Is _____ 238U{ } ^ { 238 } \mathrm { U }238U

The Binding Energy Per Nucleon for Uranium 238 Is _____ ${ } ^ { 238 } \mathrm { U }$