The mass of a ${}_{3}^{7}\mathrm{Li}$ nucleus is 0.042 u less than the sum of the masses of its nucleons. The

Question

The mass of a ${}_{3}^{7}\mathrm{Li}$ nucleus is 0.042 u less than the sum of the masses of its nucleons. The binding energy per nucleon of ${}_{3}^{7}\mathrm{Li}$ nucleus is nearly:

Accepted Answer

1. **Identify Given Data:** The mass defect ($\Delta m$) is given directly as $0.042 	ext{ u}$. This is the difference between the sum of the masses of the nucleons and the actual mass of the nucleus.
2. **Calculate Total Binding Energy (BE):** The binding energy is the energy equivalent of the mass defect. Using the standard conversion factor $1 	ext{ u} \approx 931.5 	ext{ MeV}$ :
   $$ BE = \Delta m 	imes 931.5 	ext{ MeV/u} $$
   $$ BE = 0.042 	imes 931.5 \approx 39.12 	ext{ MeV} $$
3. **Calculate Binding Energy per Nucleon:** The nucleus is Lithium-7 (${}_{3}^{7}\mathrm{Li}$), which has a mass number $A = 7$ (total number of nucleons).
   $$ BE_{	ext{per nucleon}} = \frac{BE}{A} $$
   $$ BE_{	ext{per nucleon}} = \frac{39.12 	ext{ MeV}}{7} \approx 5.58 	ext{ MeV} $$
4. **Conclusion:** The calculated value is approximately $5.6 	ext{ MeV}$, which matches Option 2.

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