The mass of the chlorine \( {}^{35}_{17}\mathrm{Cl} \) 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
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 287.82 MeV
The binding energy is the energy equivalent of the mass defect, the difference between the total mass of the separate nucleons and the mass of the nucleus. The given proton and neutron masses are the total masses of the 17 protons and the 18 neutrons of \( {}^{35}_{17}\mathrm{Cl} \).
\( \displaystyle \Delta m = (17.133025 + 18.155970) - 34.98 \)
\( \displaystyle \Delta m = 35.288995 - 34.98 = 0.308995\,\mathrm{u} \)
Using \( 1\,\mathrm{u} = 931.5\,\mathrm{MeV} \):
\( \displaystyle E_{b} = 0.308995 \times 931.5\,\mathrm{MeV} \approx 287.8\,\mathrm{MeV} \)
This is nearly 287.82 MeV.