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 (

__Answer:-__

For (^{41}_{20}Ca): Separation energy=8.363007 MeV

(^{27}_{13}Al): Separation energy=13.059 MeV

(_{0}^{1}n) is removed from a (^{41}_{20}Ca).

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

^{41}_{20}Ca --> ^{40}_{20}Ca + _{0}^{1}n

It is given that:

m (^{40}_{20}Ca) Mass = 39.962591 u,

m (^{41}_{20}Ca) Mass = 40.962278 u

m (_{0}^{1}n)Mass = 1.008665 u

The mass defect of this reaction is given as:

Therefore, ∆m= m(^{40}_{20}Ca) + (_{0}^{1}n) –m(^{41}_{20}Ca)

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

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

Therefore,∆m = 0.008978 x 931.5 (MeV/c^{2})

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

E=∆mc^{2}

=0.008978 x931.5 =8.363007MeV

For (^{27}_{13}Al), the neutron removal reaction can be written as:

^{27}_{13}Al --> ^{26}_{13}Al + _{0}^{1}n

It is given that:

m (^{27}_{13}Al)Mass = 26.981541 u

m (^{26}_{13}Al) Mass =25.986895u

The mass defect of this reaction is given as:

∆m =m (^{26}_{13}Al) +m (_{0}^{1}n) –m (^{27}_{13}Al)

=25.986895 + 1.008665 -26.981541

=0.0414019u

=0.0414019 x 931.5(MeV/c^{2})

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

E=∆mc^{2}

=(0.0414019 x 931.5) =13.059MeV

.