Class 12 Physics Nuclei Nuclear force

Nuclear force

  • The force with which the nucleons are bound together is known as nuclear force.
  • It is the strong attractive force that binds the nucleons together.
  • When the nuclear force is compared to other forces of nature like gravitational or coulomb’s force etc.it is the strongest of all the forces.
  • As protons are positively charged they repel each other.This force of repulsion is given by Coulomb’s force of repulsion.
  • This nuclear force is stronger than the coulomb’s force so it overcomes the force of repulsion.
  • This is the reason neutrons and protons are held together inside the nucleus.
  • It is independent of electric charge.Magnitude of nuclear force is same between proton-proton,proton-neutron or neutron-neutron.
  • Nuclear force cannot be given mathematically.

 

Problem: - The neutron separation energy is defined as the energy required removing a neutronfrom the nucleus. Obtain the neutron separation energies of the nuclei (4120Ca) and (2713Al) from the following data: m (4020Ca) = 39.962591 u, m (4120Ca) = 40.962278 um (2613A) = 25.986895 u and m(2713A) = 26.981541 u.

Answer:-

For (4120Ca): Separation energy=8.363007 MeV

(2713Al): Separation energy=13.059 MeV

(01n) is removed from a (4120Ca).

For a neutron nucleus. The corresponding nuclear reaction can be written as:

4120Ca --> 4020Ca + 01n

It is given that:

m (4020Ca) Mass = 39.962591 u,

m (4120Ca) Mass = 40.962278 u

m (01n)Mass = 1.008665 u

The mass defect of this reaction is given as:

Therefore, ∆m= m(4020Ca) + (01n) –m(4120Ca)

= (39.962591+1.008665-40.962278)u =0.008978u

But 1u=931.5 (MeV/c2)

Therefore,∆m = 0.008978 x 931.5 (MeV/c2)

Hence, the energy required for neutron removal is calculated as:

E=∆mc2

=0.008978 x931.5 =8.363007MeV

For (2713Al), the neutron removal reaction can be written as:

2713Al --> 2613Al + 01n

It is given that:

m (2713Al)Mass = 26.981541 u

m (2613Al) Mass =25.986895u

The mass defect of this reaction is given as:

∆m =m (2613Al) +m (01n) –m (2713Al)

=25.986895 + 1.008665 -26.981541

=0.0414019u

=0.0414019 x 931.5(MeV/c2)

Hence, the energy required for neutron removal is calculated as:

E=∆mc2

=(0.0414019 x 931.5) =13.059MeV

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