JEE Main 2023 · 1 February · evening · Numerical answer · +4 / −0
Nucleus A having Z=17 and equal number of protons and neutrons has 1.2 MeV binding energy per nucleon.
Another nucleus B of Z=12 has total 26 nucleons and 1.8 MeV binding energy per nucleons.
The difference of binding energy of B and A will be _____________ MeV.
Answer and solution
Answer 6
Solution
Physics sir · 3 steps
Binding energy per nucleon times the number of nucleons gives the total. Find both totals, then subtract.
- Nucleons in A. A has Z = 17 protons and the same number of neutrons, so its mass number is 17 + 17 = 34.
Total BE = A × (BE/nucleon)
- Total for each. For A: 34 × 1.2 = 40.8 MeV. For B: 26 × 1.8 = 46.8 MeV. The proton number Z = 12 of B plays no part, because the nucleon count is already given.
- So. 46.8 - 40.8 = 6 MeV, so the difference is 6.
The common slip: 0.6. Subtracting the per-nucleon values, 1.8 - 1.2, gives 0.6. The question asks for the difference in total binding energy.
The idea to keep. Per-nucleon binding energy measures how stable a nucleus is. The total binding energy is that value times the mass number, so always convert to totals before you compare two nuclei.
Answer from the NTA final answer key. Solution by PrepWiser.