Question

Find the Binding energy per neucleon for \[\ce{^120_50Sn}\]. Mass of proton m_p = 1.00783 U, mass of neutron m_n = 1.00867 U and mass of tin nucleus m_Sn = 119.902199 U. (take 1 U = 931 MeV)

विकल्प

• 7.5 MeV

• 9.0 MeV

• 8.0 MeV

• 8.5 MeV

Answer 1

8.5 MeV

Explanation:

Mass defect,

Δm = (50m_p + 70m_n) - (m_sn)

= (50 × 1.00783 + 70 × 1.008) - (119.902199) = 1.096

Binding energy = (Δm)C² = (Δm) × 931 = 1020.56

`"Binding energy"/"Nucleon" = 1020.5631/120`

= 8.5 MeV