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      mn: mass of a neutron;

mp: mass of a proton

        M (Z, A): mass of bounded nucleus

Then, Δm = Z . mp + (A – Z). mn – 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

1884_nucleons.JPG

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:

                2387_equations.JPG

Given that

            mn = 1.6747 × 10–27 kg  

            mp = 1.6725 × 10–27 kg  

            me = 0.00091 × 10–27 kg  

        (A)    0.931 MeV                 (B)    0.731 MeV

        (C)    0.511 MeV                 (D)    0.271 MeV

Solution:

Mass defect Δm = (1.6747 – 1.6725 – 0.0091) × 10–27 

                              = 0.0012 × 10–27 kg 

        ΔE = 0.0012 × 10–27 × (3 × 108)2 / 1.6 × 10–12 = 0.731 MeV 

 

Illustration 5:

The binding energy per nucleon for 3Li7 will be, if the mass of 3Li7 is 7.01653 amu.

        (A)    5.6 MeV                     (B)    39.25 MeV

        (C)    1 MeV                        (D)    zero.

Solution:

              E = ΔE / A = Δm × 931 / A MeV  

            Δm = (3mp + 4mn) – mass of Li7 

            = (3 × 1.00759 + 4 × 1.008898) – 7.01653

            = 0.04216

            ΔE = 0.04216 × 931 / 7 = 39.25 / 7 = 5 MeV

Illustration 6:

How much energy is released in the following reaction?

                        1H2 + 1H2 = 2He4 

If the B.E./Nucleon of 1H2 and 2He4 are 1.123 MeV and 7.2 MeV respectively.

        (A)    12 MeV                      (B)    24.3 MeV

        (C)    36 MeV                      (D)    zero

Solution:

B. E. of 1H2

          ΔE = 1.125

        E = A × ΔE

        E = 2 × 1.125

           = 2.25 MeV

         B.E. of two 1H2 = 2.25

         Ed = 4.5 MeV 

        B.E. of an α -particle = 4 × 7.2

                                   Ea = 28.8

        Energy released ER = Ea – Ed

                                   ER = 28.8 – 4.5 = 24.3 MeV
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