The neutral $^{27}_{13}$Al atom has a mass of 26.981539 u; a neutral hydrogen atom has a mass of 1.007825 u; a neutron has a mass of 1.008665 u; and a proton has a mass of 1.007277 u. What is the binding energy per nucleon for $^{27}_{13}$Al? (1 u = 931.5 MeV/c²)

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Quizplus · Accepted Answer

To find the binding energy per nucleon, first calculate the mass defect:  $$ \text{Mass of protons} = 13 \times 1.007277 \, \text{u} $$  $$ \text{Mass of neutrons} = 14 \times 1.008665 \, \text{u} $$  $$ \text{Total mass of nucleons} = 13 \times 1.007277 + 14 \times 1.008665 $$  $$ \text{Mass defect} = (\text{Total mass of nucleons}) - 26.981539 \, \text{u} $$  Convert the mass defect to energy using $1 \, \text{u} = 931.5 \, \text{MeV/c}^2$.  Finally, divide by the number of nucleons (27) to find the binding energy per nucleon.  The calculation gives approximately 8.3 MeV per nucleon.

The Neutral 1327^{27}_{13}1327​ Al Atom Has a Mass of 26 1327^{27}_{13}1327​ Al? (1 U = 931

The Neutral $^{27}_{13}$ Al Atom Has a Mass of 26 $^{27}_{13}$ Al? (1 U = 931