Question

The mass of proton is 1.0073 u and that of neutron is 1.0087 u (u = atomic mass unit). The binding energy of 42He is (Given : helium nucleus mass ≈ 4.0015 u)

Answer 1

Δm = (2 × 1.0074 + 2 × 1.0087 - 4.0015) = 0.0307

E = (Δm) × 931 Mev = 0.0307 × 931 = 28.5 Mev

Answer 2

Answer:

28.4 Mev

Explanation:

Binding energy = 42He (Given)

Mass of helium nucleus ≈ 4.0015 u (Given)

Mass of proton = 1.0073 u (Given)

Mass of neutron = 1.0087 u (Given)

where u is the atomic mass unit

Thus, Mass defect = 2Mp + 2Mn - Mhe

Therefore, following the equation -

Δm = (2 × 1.0073 + 2 × 1.0087 - 4.0015) = 0.0305

Binding energy = ( 931 ×  mass defect) MeV

E = (Δm) × 931 Mev  (1 amu = 931 Mev).

= 0.0305 × 931

= 28.4 MeV

Thus, the binding energy of 42He is 28.4 MeV.