Mass Defect Binding Energy

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 = Δm . c²

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

From the plot of B.E./nucleons Vs mass number (A), we observe that:

(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.

Illustration 2:

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 

Illustration 4:

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

Given that