Question
A nucleus of mass number A, has a mass defect $\triangle \text{m}.$ Give the formula, for the binding energy per nucleon, of this nucleus.
✓
Answer
BE per nucleon, Bn $=\frac{\text{Total binding energy}}{\text{Number of nucleons}}$
$=\frac{\triangle \text{mc}^2}{\text{A}}$
Where c is the speed of light in vacuum.
$=\frac{\triangle \text{mc}^2}{\text{A}}$
Where c is the speed of light in vacuum.
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