Mass Defect & Binding energy

Mass defect :

The actual mass of a nucleus is experimentally observed to be smaller than the sum of the masses of free nucleons constituting it.

The difference between the experimental mass , m(ZXA) , of a nucleus and the sum of the masses of free nucleons (Z protons, and (A−Z) neutrons) is known as the mass defect .

In a nucleus , ZXA

Mass Defect is

Δm = Zmp + (A-Z) mn − m (ZXA)

What is Binding Energy ?

The amount of energy needed to disintegrate a nucleus into its constituent nucleons is called the binding energy. It is the energy equivalent to the mass defect.

B.E. = Δmc2 , where c = velocity of light.

If mass is measured in amu then B.E. = Δm × 931.5 MeV .

As 1 amu = 931.5 MeV/c2.

B.E. per nucleon = Binding fraction

$\large \frac{Binding \; Energy}{Mass \; number} = \frac{\Delta m \times 931.5 MeV}{A}$

Packing Fraction of a Nucleus:

Packing fraction: It is defined as the mass defect per nucleon.

$\large Packing \; fraction = \frac{\Delta m }{A}$

Packing fraction measures the stability of a nucleus. Smaller the value of packing fraction, large is the stability of the nucleus.

Illustration : The binding energy of 17Cl35 nucleus is 298 MeV. Find its atomic mass. The mass of hydrogen atom (1H1) is 1.008143 a.m.u. and that of a neutron is 1.008986 a.m.u.
Given 1 a.m.u. = 931 MeV.

Solution: The 17Cl35 atom has 17 protons and 18 neutrons in its nucleus.

Mass of 17 1H1 atom has 17 protons and 18 neutrons in its nucleus.

Mass of 17 protons = 17 × 1.008143 amu

= 17.138431 amu.

Mass of 18 neutrons = 18 × 1.008986

= 18.161748 a.m.u.

Total = 35.300179 a.m.u

Mass defect Δm = 298/931

= 0.320085 a.m.u.

The atomic mass of 17Cl35 would be the sum of equivalent of the binding energy of the nucleus.

Hence atomic mass of
17Cl35 = 35.300179 − 0.320085

= 34.980094 a.m.u

Also Read :

Size of the nucleus
Nuclear Stability (Nuclear force)
Q-Value
Problem Solving technique (In nuclear physics)

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