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 Ebn.
  • Experimentally there was a graph plotted between binding energy per nucleon and the mass number(A).

 

  • Following are the observations from the graph:-
  1. Initially the graph was increasing.This implies that Ebn is very less for lesser mass number.
  2. In the middle range the Ebn becomes constant.This means Ebn is independent of mass number.
  3. In the end Ebnstarts decreasing.This shows Ebn is less when mass number is more.

Problem: - Obtain the binding energy (in MeV) of a nitrogen nucleus (147N), given m (147N) =14.00307u?

Answer:- Atomic mass of (147N ) nitrogen, m = 14.00307 u

A nucleus of (147N) nitrogen contains 7 protons and 7 neutrons.

Hence, the mass defect of this nucleus, ∆m = 7mH + 7mn − m

Where,

Mass of a proton, mH = 1.007825 u

Mass of a neutron, mn= 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/c2

∆m = 0.11236 × 931.5 MeV/c2

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

Eb = ∆mc2

Where, c = Speed of light

Eb = 0.11236 × 931.5(MeV/c2)/c2

= 104.66334 MeV

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

Deriving Nuclear force from Ebn

  1. Lighter nuclei: -
    1. In the initial part of the graph A(mass number) is less therefore Ebn is also less.As a result lesser energy is required to separate the nucleons.
    2. This showsnuclei areunstable.
    3. The nuclei are unstable and in order to become stable lighter nuclei combine with each other to form heavier nuclei.
    4. Let the energy of heavier nuclei formed is E’bn and of lighter nuclei be Ebn. This implies E’bn> Ebn.
    5. Energy is released when 2 lighter nuclei combine together to form a heavier nuclei.
    6. This process is known as Nuclear Fusion.
  2. For heavier nuclei:-
    1. Mass number is very high and Ebn is very less.
    2. In order to become stable the heavier nuclei will split into 2 lighter nuclei.
    3. Energy associated with heavier nuclei =Ebn and energy associated with 2 lighter nuclei =E’bn.
    4. This implies E’bn> Ebn. Energy is released in this process by the heavier nuclei in order to attain stability.
    5. This process is known as Nuclear Fission.
  3. Constancy of Ebnin the mid-range of A:-
    1. In this portion the mass number is increasing due to whichnumber of nucleons also increase.
    2. The force which is present between the nucleons is of short range.The strength of the force decreases as the distance increases.
    3. The nucleons are getting affected by their nearest neighbouring nucleons and not by the nucleons which are far away.
    4. As a result Ebn remains constant.
    5. But when there are too many nucleons Ebn suddenly starts decreasing.

Problem: - Obtain the binding energy of the nuclei (5626Fe) and (20983Bi) in units of MeV from the following data: m (5626Fe) = 55.934939u and m (20983Bi) = 208.980388 u.

Answer:- Atomic mass of (5626Fe), mFe = 55.934939 u

(5626Fe) nucleus has 26 protons and (56 − 26) = 30 neutrons

Hence, the mass defect of the nucleus, ∆m = ((26 × mH) + (30 × mn)– mFe)

Where,

Mass of a proton, mH = 1.007825 u

Mass of a neutron, mn = 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/c2

Therefore, ∆m = 0.528461 × 931.5 MeV/c2

The binding energy of this nucleus is given as:

Eb1 = ∆mc2

Where, c =Speed of light

Eb1 = 0.528461 × 931.5(MeV/c2)/c2

= 492.26 MeV

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

Atomic mass of (20983Bi), m2 = 208.980388 u

(20983Bi) nucleus has 83 protons and (209 − 83) = 126 neutrons.

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

∆m' = (83 × mH + 126 × mn) − m2

Where,

Mass of a proton, mH = 1.007825 u

Mass of a neutron, mn = 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/c2

Therefore, ∆m' = 1.760877 × 931.5 (MeV/c2) x c2

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

Eb2 = ∆m'c2

= 1.760877 × 931.5(MeV/c2) x c2

= 1640.26 MeV

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

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