Find the Binding energy per neucleon for ₅₀¹²⁰Sn. Mass of proton mₚ = 1.00783 U, mass of neutron mₙ = 1.00867 U and mass of tin nucleus m Sn = 119.902199 U. (take 1U = 931 MeV)

Solution

Physics sir · 4 steps

Weigh the 120 loose nucleons, subtract the nucleus, convert the missing mass to MeV, then share it among 120.

1. Count the nucleons. For ¹²⁰₅₀Sn, Z = 50 protons and N = A - Z = 120 - 50 = 70 neutrons.
2. Mass of free nucleons. 50 × 1.00783 = 50.3915 u and 70 × 1.00867 = 70.6069 u. Together they weigh 120.9984 u.

   Δ m = 120.9984 - 119.902199 = 1.096201 u

3. Convert to energy. Total binding energy = 1.096201 × 931 ≈ 1020.6 MeV. Dividing by the 120 nucleons gives 1020.6/120 ≈ 8.50 MeV per nucleon.
4. So. The binding energy per nucleon of tin-120 is about 8.5 MeV, which is option (b).

The idea to keep. Binding energy per nucleon is the mass defect times 931 MeV, divided by A. Divide by the mass number A, never by Z or N. Most stable nuclei come out between 7.5 and 8.8 MeV.

Answer from the NTA final answer key. Solution by PrepWiser.