Question

The mass of the proton is 1.0073 u and that of the neutron is 1.0087 u ( u = atomic mass unit). The binding energy of ${}_2^{4}He$ is :

Given, mass of helium nucleus ≈ 4.0015u

• A. 0.0305 J
• B. 0.0305 erg
• C. 28.4 MeV
• D. 0.061 u

Answer 1

Answer: (C)

28.4 MeV

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 - M(He)

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 He is 28.4 MeV.

Option C is correct.