Class 12 Physics Nuclei | Nuclear binding energy per nucleon |

**Nuclear binding energy per nucleon**

- Nuclear binding energy per nucleon is defined as the average energy per nucleon needed to separate a nucleus into its individual constituents.
- It is denoted by E
_{bn}. - Experimentally there was a graph plotted between binding energy per nucleon and the mass number(A).

- Following are the observations from the graph:-

- Initially the graph was increasing.This implies that E
_{bn}is very less for lesser mass number. - In the middle range the E
_{bn}becomes constant.This means E_{bn}is independent of mass number. - In the end E
_{bn}starts decreasing.This shows E_{bn}is less when mass number is more.

__Problem__**: -** Obtain the binding energy (in MeV) of a nitrogen nucleus (^{14}_{7}N), given m (^{14}_{7}N) =14.00307u?

** Answer:- **Atomic mass of (

A nucleus of (^{14}_{7}N) nitrogen contains 7 protons and 7 neutrons.

Hence, the mass defect of this nucleus, ∆m = 7m_{H} + 7m_{n} − m

Where,

Mass of a proton, m_{H} = 1.007825 u

Mass of a neutron, m_{n}= 1.008665 u

Therefore, ∆m = (7 × 1.007825 + 7 × 1.008665 − 14.00307)

= 7.054775 + 7.06055 − 14.00307

= 0.11236 u

But 1 u = 931.5 MeV/c^{2}

∆m = 0.11236 × 931.5 MeV/c^{2}

Hence, the binding energy of the nucleus is given as:

E_{b} = ∆mc^{2}

Where, c = Speed of light

E_{b} = 0.11236 × 931.5(MeV/c^{2})/c^{2}

= 104.66334 MeV

Hence, the binding energy of a nitrogen nucleus is 104.66334 MeV.

__Deriving Nuclear force from E _{bn}__

- Lighter nuclei: -
- In the initial part of the graph A(mass number) is less therefore E
_{bn}is also less.As a result lesser energy is required to separate the nucleons. - This showsnuclei areunstable.
- The nuclei are unstable and in order to become stable lighter nuclei combine with each other to form heavier nuclei.
- Let the energy of heavier nuclei formed is E’
_{bn}and of lighter nuclei be E_{bn}. This implies E’_{bn}> E_{bn}. - Energy is released when 2 lighter nuclei combine together to form a heavier nuclei.
- This process is known as
__Nuclear Fusion__.

- In the initial part of the graph A(mass number) is less therefore E
- For heavier nuclei:-
- Mass number is very high and E
_{bn}is very less. - In order to become stable the heavier nuclei will split into 2 lighter nuclei.
- Energy associated with heavier nuclei =E
_{bn}and energy associated with 2 lighter nuclei =E’_{bn}. - This implies E’
_{bn}> E_{bn}. Energy is released in this process by the heavier nuclei in order to attain stability. - This process is known as
__Nuclear Fission__.

- Mass number is very high and E
- Constancy of E
_{bn}in the mid-range of A:-- In this portion the mass number is increasing due to whichnumber of nucleons also increase.
- The force which is present between the nucleons is of short range.The strength of the force decreases as the distance increases.
- The nucleons are getting affected by their nearest neighbouring nucleons and not by the nucleons which are far away.
- As a result E
_{bn}remains constant. - But when there are too many nucleons E
_{bn}suddenly starts decreasing.

** Problem: -** Obtain the binding energy of the nuclei (

** Answer:- **Atomic mass of (

(^{56}_{26}Fe) nucleus has 26 protons and (56 − 26) = 30 neutrons

Hence, the mass defect of the nucleus, ∆m = ((26 × m_{H}) + (30 × m_{n})– m_{Fe})

Where,

Mass of a proton, m_{H} = 1.007825 u

Mass of a neutron, m_{n} = 1.008665 u

∆m = ((26 × 1.007825 + 30 × 1.008665) − 55.934939)

= (26.20345 + 30.25995 − 55.934939)

= 0.528461 u

But 1 u = 931.5 MeV/c^{2}

Therefore, ∆m = 0.528461 × 931.5 MeV/c^{2}

The binding energy of this nucleus is given as:

E_{b1} = ∆mc^{2}

Where, c =Speed of light

E_{b1} = 0.528461 × 931.5(MeV/c^{2})/c^{2}

= 492.26 MeV

Average binding energy per nucleon= (492.26/56)=8.76 MeV

Atomic mass of (^{209}_{83}Bi), m_{2} = 208.980388 u

(^{209}_{83}Bi) nucleus has 83 protons and (209 − 83) = 126 neutrons.

Hence, the mass defect of this nucleus is given as:

∆m' = (83 × m_{H} + 126 × m_{n}) − m_{2}

Where,

Mass of a proton, m_{H} = 1.007825 u

Mass of a neutron, m_{n} = 1.008665 u

∆m' = (83 × 1.007825 + 126 × 1.008665 – 208.980388)

= 83.649475 + 127.091790 − 208.980388

= 1.760877 u

But 1 u = 931.5 MeV/c^{2}

Therefore, ∆m' = 1.760877 × 931.5 (MeV/c^{2}) x c^{2}

Hence, the binding energy of this nucleus is given as:

E_{b2} = ∆m'c^{2}

= 1.760877 × 931.5(MeV/c^{2}) x c^{2}

= 1640.26 MeV

Average binding energy per nucleon = (1640.26/209) =7.848MeV

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