Mass Defect and Binding Energy

The rest mass of the stable nucleus of a stable atom is always less than the sum of the masses of constituent nucleons. The difference is called the mass defect Δm (i.e., Δm.c²) is utilised in keeping the nucleons bound together. This energy is known as the binding energy. In order to break the nucleus into its constituent nucleons an amount of energy equal to its binding energy has to be supplied to the nucleus. The mass defect per nucleon Δm/A = P, is called the packing fraction of the nucleus.

Atomic mass is the mass of a single atomic particle or molecule. It is the sum of protons and electrons present in the atom of an element. It is expressed in mole. It is simply a collection of nuclides that make up a chemical element. it is a whole number. 

Atomic weight is the ratio of atom of an element. The average weight of an atom is relative to the 1/12 weight of the carbon -12 atom. It is also referred to as relative atomic mass.The value is not necessarily a whole number.

Atomic weight = [Mass(a)isotope(a)]+[Mass(b)isotope(b)]

The nucleons are bound together in a nucleus and the energy has to be supplied in order to break apart the constituents into free nucleons. The energy with which nucleons are bounded together in a nucleus is called as Binding Energy (B.E.). In order to free nucleons from a bounded nucleus this much of energy (= B.E.) is to be supplied.

It is observed that the mass of a nucleus is always less than the mass of constituent (free) nucleons. This difference in mass is called as mass defect and is denoted as Dm.

If m_n: mass of a neutron;

m_p: mass of a proton

M (Z, A): mass of bounded nucleus

Then, Δm = Z . m_p + (A – Z). m_n – M (Z, A)

This mass-defect is in form of energy and is responsible for binding the nucleons together. From Einstein's law of inter-conversion of mass into energy:

E = mc²     (c: speed of light; m: mass)

Binding energy,

Generally, Δm is measured in amu units. So let us calculate the energy equivalent to 1 amu. It is calculated in eV (electron volts; 1 eV =1.6 x10^–19J)

E (= 1 amu = 1.67 × 10^–27 (3×10⁸)² / 1.6 × 10^–19) eV = 931 × 10⁸ eV = 931 MeV

=> B.E. = Δm (931) MeV

There is another quantity which is very useful in predicting the stability of a nucleus called as Binding energy per nucleons.

B.E. per nucleons = Δm (931) / A MeV

Observation from the plot of B.E./nucleons Vs mass number (A):-

(i) B.E./nucleons increases on an average and reaches a maximum of about 8.7 MeV for A º 50 – 80.

(ii) For more heavy nuclei, B.E./nucleons decreases slowly as A increases. For the heaviest natural element U²³⁸ it drops to about 7.5 MeV.

(iii) From above observation, it follows that nuclei in the region of atomic masses 50-80 are most stable.

Refer this video for better understanding about mass defect and binding energy:-

Problem 1:-:

If mass of proton = 1.008 amu and mass of neutron = 1.009 amu, then the binding energy per nucleon for ₄Be⁹ (mass = 9.012 amu) will be:

(A) 0.0672 MeV                (B) 0.672 MeV

(C) 6.72 MeV                   (D) 67.2 MeV

Solution:-

Mass defect,

Δm = (4 × 1.008 + 5 × 1.009) – 9.012

= 9.077 – 9.012 =0.065 amu

BE/A = 0.065 × 931 / 9 = 6.72 MeV 

Problem 2:-:

The energy released in the following b-decay process will be:

Given that,

m_n = 1.6747 × 10^–27 kg  

m_p = 1.6725 × 10^–27 kg

m_e = 0.00091 × 10^–27 kg  

(A)    0.931 MeV                 (B)    0.731 MeV

(C)    0.511 MeV                 (D)    0.271 MeV