Calculate:
i) Mass defect of helium   and 
ii) 

Its binding energy in MeV.

Mass of a proton = 1.007276 u

Mass of a neutron = 1.008665 u
Mass of  nucleus = 4.001506 u

An α-particle is helium (₂He⁴) nucleus, containing two protons and two neutrons.

We can find out its mass defect from the given data:

Mass of 2 protons = 2 x 1.000728

                       = 2.01456 u

Mass of 2 neutrons = 2 x 1.00867

                         = 2.01734u

Total = 4.03190 u.

This is the total mass of the nucleus of α-particle.

The mass of α-particle is 4.00151 u. therefore, the mass defect is

Δm = mass of nucleons-mass of α-particle

     = 4.03190 - 4.00151u

     = 0.03039 u.

According to the energy mass relation ΔE = (Δm)c², 1u of mass equivqlent to 931.5 MeV of energy.

Energy equivalent to 0.03039 u is

ΔE = 0.03039X 931.5 = 28.3 MeV.

This is the binding energy of α-particle.

The binding energy per nucleon is 28.3/4 = 7.07Mev.

This large amount of energy explains the high stability of a- particle.