Nuclear Forces

The protons and neutrons are held together by the strong attractive forces inside the nucleus. These forces are called as nuclear forces.

(i) Nuclear forces are short-ranged. They exist in small region (of diameter 10–15 m = 1 fm). The nuclear force between two nucleons decrease rapidly as the separation between them increases and becomes negligible at separation more than 10 fm.

(ii) Nuclear force are much stronger than electromagnetic force or gravitational attractive forces.

(iii) Nuclear force are independent of charge. The nuclear force between two proton is same as that between two neutrons or between a neutron and proton. This is known as charge independent character of nuclear forces.

In a typical nuclear reaction

(i) In nuclear reactions, sum of masses before reaction is greater than the sum of masses after the reaction. The difference in masses appears in form of energy following the Law of inter-conversion of mass & energy. The energy released in a nuclear reaction is called as Q Value of a reaction and is given as follows.

If difference in mass before and after the reaction is Δm amu (Δm = mass of reactants minus mass of products) then Q value = Δm (931) MeV

(ii) Law of conservation of momentum is also followed.

(iii) Total number of protons and neutrons should also remain same on both sides of a nuclear reaction.

Illustration 7:

Calculate the Q-value of the nuclear reaction:

                0C12  10Ne20 + 2He4

The following data are given:

                m(6C12) = 12.000000 u

                m(10Ne20) = 12.000000 u

                m(2Ne4) = 4.002603 u

Solution:

The Q-value of this reaction may be easily calculated may be easily calculated from the masses of the individual nuclei.

                   Q = [2m (6C12) – {m (10Ne20) + m (2He4)}]e2

                   = 24.000000 – (19.992439 + 4.002603)} u × c2

                   = 4.618 MeV

Illustration 8:

Binding energy of the calculate the mass defect and the binding energy of an α-particle given the following date:

mn = 1.0086665 u

mp = 1.007825 u

mα = 4.0026 u

(1u = 931.5 MeV/c2)

Solution:

The mass defect of an α-particle is given by

          Δm = (2 × 1.008665 + 2 × 1.007825) – 4.0026)u

                = 0.03038u

The Binding energy is related to the mass defect by the reaction.

               B.E. = Δm.c2 

                     = 0.03038 × 931.5 MeV

                     = 28.3 MeV (approximately)

NUCLEAR FORCES

The protons and neutrons are held together by the strong attractive forces inside the nucleus. These forces are called as nuclear forces.

(i) Nuclear forces are short-ranged. They exist in small region (of diameter 10–15 m = 1 fm). The nuclear force between two nucleons decrease rapidly as the separation between them increases and becomes negligible at separation more than 10 fm.

(ii) Nuclear force are much stronger than electromagnetic force or gravitational attractive forces.

(iii) Nuclear force are independent of charge. The nuclear force between two proton is same as that between two neutrons or between a neutron and proton. This is known as charge independent character of nuclear forces.

In a typical nuclear reaction

(i) In nuclear reactions, sum of masses before reaction is greater than the sum of masses after the reaction. The difference in masses appears in form of energy following the Law of inter-conversion of mass & energy. The energy released in a nuclear reaction is called as Q Value of a reaction and is given as follows.

If difference in mass before and after the reaction is Δm amu

m = mass of reactants minus mass of products)

then Q value = Δm (931) MeV

(ii) Law of conservation of momentum is also followed.

(iii) Total number of protons and neutrons should also remain same on both sides of a nuclear reaction.

Illustration 7:

Calculate the Q-value of the nuclear reaction:

                0C12  10Ne20 + 2He4

The following data are given:

                m(6C12) = 12.000000 u

                m(10Ne20) = 12.000000 u

                m(2Ne4) = 4.002603 u

Solution:

The Q-value of this reaction may be easily calculated may be easily calculated from the masses of the individual nuclei.

                   Q = [2m (6C12) – {m (10Ne20) + m (2He4)}]e2

                   = 24.000000 – (19.992439 + 4.002603)} u × c2

                   = 4.618 MeV

Illustration 8:

Binding energy of the calculate the mass defect and the binding energy of an α-particle given the following date:

mn = 1.0086665 u

mp = 1.007825 u

mα = 4.0026 u

(1u = 931.5 MeV/c2)

Solution:

The mass defect of an α-particle is given by

          Δm = (2 × 1.008665 + 2 × 1.007825) – 4.0026)u

                = 0.03038u

The Binding energy is related to the mass defect by the reaction.

               B.E. = Δm.c2 

                     = 0.03038 × 931.5 MeV

                     = 28.3 MeV (approximately)

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