Answer

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**Hint:**When a nucleus is stable, energy is required from outside to disrupt it into its constituents separately, whereas when a nucleus is unstable, it will disintegrate by itself. So binding energy of a nucleus gives us information about its stability.

Formulas used:

$ {E_B} = \left[ {Z{m_p} + \left( {A - Z} \right){m_n} - {M_n}} \right]{c^2} $ Where $ {m_p} $ is the mass of a proton and $ {m_n} $ is the mass of a neutron both equal to $ 1.67 \times {10^{ - 27}}kg $

$ {M_n} $ is the mass of a nucleus, $ c $ is the speed of light in vacuum, $ A $ is the mass number and $ Z $ is the atomic number.

$ {f_B} = \dfrac{{{E_B}}}{A} = \left[ {\dfrac{{Z{m_p} + \left( {A - Z} \right){m_n} - {M_n}}}{A}} \right]{c^2} $

Where $ {f_B} $ is the binding energy per nucleon.

**Complete step by step answer**

Binding energy of a nucleus is defined as the minimum amount of energy which must be supplied to the nucleus to break it into its constituent nucleons. Now, the specific binding energy is defined as the average binding energy per nucleon.

The mass of any stable nucleus is found to be less than the sum of the mass of all the protons and the neutrons inside the nucleus. This lost mass in the form of energy keeps the nucleus together and is termed as binding energy.

If $ Z $ is the atomic number and $ A $ is the mass number then the binding energy is given as,

$ {E_B} = \left[ {Z{m_p} + \left( {A - Z} \right){m_n} - {M_n}} \right]{c^2} $

Where $ {m_p} $ is the mass of a proton and $ {m_n} $ is the mass of a neutron both equal to $ 1.67 \times {10^{ - 27}}kg $

$ {M_n} $ is the mass of a nucleus and $ c $ is the speed of light in vacuum.

The expression for the binding energy per nucleon or the specific binding energy is given as

$ {f_B} = \dfrac{{{E_B}}}{A} = \left[ {\dfrac{{Z{m_p} + \left( {A - Z} \right){m_n} - {M_n}}}{A}} \right]{c^2} $

Stability and binding energy :- The binding energy of a nucleus gives us a measure of its stability.

-If the binding energy is positive, then the nucleus is stable.

-If the binding energy is negative, the nucleus is unstable and will disintegrate itself.

The diagram below shows us the variation of binding energy per nucleon with mass number $ A $

From this diagram we can tell that the elements with the most stable nucleus have mass number between 40 and 60. The elements below this range undergo fusion and the elements above this range undergo fission.

**Note**

As mass is expressed in $ a.m.u $ (atomic mass unit), in the unit of energy ( $ 1a.m.u = 931.5MeV $ )

1. Binding energy can be expressed as, $ {E_B} = \left[ {Z{m_p} + \left( {A - Z} \right){m_n} - {M_n}} \right] \times 931.5MeV $

2. Binding energy per nucleon is given as, $ \dfrac{{{E_B}}}{A} = \dfrac{{Z{m_p} + \left( {A - Z} \right){m_n} - {M_n}}}{A} \times 931.5MeV $ per nucleon.

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