The mass of the chlorine ${^{}_{17}Cl}^{35}$ nucleus is 34.98 amu, which has protons of mass 17.133025 amu and neutrons of mass 18.155970 amu, The binding energy of the nucleus is nearly

287.82 MeV

The correct answer is Option (1) → 287.82 MeV

Given:

Mass of $^{35}_{17}\text{Cl}$ nucleus, $M_\text{nucleus} = 34.98\ \text{amu}$

Mass of protons: $17 \times 1.007825 = 17.133025\ \text{amu}$

Mass of neutrons: $18 \times 1.008665 = 18.155970\ \text{amu}$

Total mass of nucleons: $M_\text{nucleons} = 17.133025 + 18.155970 = 35.288995\ \text{amu}$

Mass defect: $\Delta m = M_\text{nucleons} - M_\text{nucleus} = 35.288995 - 34.98 = 0.308995\ \text{amu}$

Binding energy: $E_b = \Delta m \cdot 931.5\ \text{MeV/amu} \approx 0.309 \cdot 931.5 \approx 288\ \text{MeV}$

∴ Binding energy of $^{35}_{17}\text{Cl}$ nucleus ≈ 288 MeV