To solve this problem, we need the following information, which is partially missing from the image: 1. The specific isotope of lithium (number of nucleons). 2. The mass of a neutron. 3. The actual atomic mass of the lithium isotope. I will assume the most common isotope, Lithium-7 ($^7_3\text{Li}$), and use standard values for the neutron mass and the atomic mass of Lithium-7. Assumptions: Lithium isotope is Lithium-7 ($^7_3\text{Li}$). Number of protons ($Z$) = 3 Number of neutrons ($N$) = $7 - 3 = 4$ Total number of nucleons ($A$) = 7 Mass of proton ($m_p$) = $1.00728 \, \text{a.m.u.}$ (given) Mass of neutron ($m_n$) = $1.00866 \, \text{a.m.u.}$ (standard value) Atomic mass of Lithium-7 ($m_{\text{Li-7}}$) = $7.016003 \, \text{a.m.u.}$ (standard value) Energy equivalent of $1 \, \text{a.m.u.}$ = $931.5 \, \text{MeV}$ Step 1: Calculate the total mass of the individual constituent particles (protons and neutrons). Mass of 3 protons ($3 \times m_p$): $$3 \times 1.00728 \, \text{a.m.u.} = 3.02184 \, \text{a.m.u.}$$ Mass of 4 neutrons ($4 \times m_n$): $$4 \times 1.00866 \, \text{a.m.u.} = 4.03464 \, \text{a.m.u.}$$ Total mass of constituent particles ($m_{\text{constituents}}$): $$m_{\text{constituents}} = 3.02184 \, \text{a.m.u.} + 4.03464 \, \text{a.m.u.}$$ $$m_{\text{constituents}} = 7.05648 \, \text{a.m.u.}$$ Step 2: Calculate the mass defect ($\Delta m$). The mass defect is the difference between the total mass of the individual nucleons and the actual atomic mass of the nucleus. $$\Delta m = m_{\text{constituents}} - m_{\text{Li-7}}$$ $$\Delta m = 7.05648 \, \text{a.m.u.} - 7.016003 \, \text{a.m.u.}$$ $$\Delta m = 0.040477 \, \text{a.m.u.}$$ Step 3: Calculate the nuclear binding energy ($E_b$). The binding energy is calculated using the mass defect and the energy equivalent of 1 a.m.u. $$E_b = \Delta m \times 931.5 \, \text{MeV/a.m.u.}$$ $$E_b = 0.040477 \, \text{a.m.u.} \times 931.5 \, \text{MeV/a.m.u.}$$ $$E_b = 37.7039555 \, \text{MeV}$$ Rounding to a reasonable number of significant figures (e.g., 5, based on the input masses): $$E_b \approx 37.704 \, \text{MeV}$$ Step 4: Calculate the average nuclear binding energy per nucleon ($E_{b,avg}$). The average binding energy per nucleon is the total binding energy divided by the total number of nucleons ($A$). $$E_{b,avg} = \frac{E_b}{A}$$ $$E_{b,avg} = \frac{37.7039555 \, \text{MeV}}{7}$$ $$E_{b,avg} = 5.386279357 \, \text{MeV/nucleon}$$ Rounding to a reasonable number of significant figures: $$E_{b,avg} \approx 5.3863 \, \text{MeV/nucleon}$$ The nuclear binding energy for Lithium-7 is $\boxed{\text{37.704 MeV}}$. The average nuclear binding energy per nucleon for Lithium-7 is $\boxed{\text{5.3863 MeV/nucleon}}$.