Class 12 Physics Nuclei | Nuclear Energy |

**Nuclear Energy**

- Nuclear energy is the energy that holds together the nuclei of atoms.
- Nuclear energy is obtained from nucleusby either:-
- Breaking of heavy nucleus into 2 relatively lighter nuclei known as
__nuclear fission__or by - Combining 2 lighter nuclei to form a heavy nucleus known as
__nuclear fusion__.

- Breaking of heavy nucleus into 2 relatively lighter nuclei known as
- Nuclear energy is becominga possible solutionfor the energy crisis in the world.
- Electric energy can be harnessed from nuclear energy.

Nuclear reactor

__Types of Nuclear reactions__

** Problem:- **A given coin has a mass of 3.0 g. Calculate the nuclear energy that would be required to separate all the neutrons and protons from each other. For simplicity assume that thecoin is entirely made of (

** Answer:**- Mass of a copper coin, m’ = 3 g

Atomic mass of (_{29}^{63}Cu) atom, m = 62.92960 u

The total number of (_{29}^{63}Cu) atoms in the coin, N = (N_{A} x m’)/ (Mass number)

Where,

N_{A} = Avogadro’s number = 6.023 × 10^{23} atoms /g

Mass number = 63 g

Therefore N = (6.023 × 10^{23} x3)/ (63) =2.868x10^{22} atoms.

(_{29}^{63}Cu) nucleus has 29 protons and (63-29) = 34 neutrons.

Therefore, Mass defect of this nucleus, ∆m' = 29 × m_{H} + 34 × m_{n} − m

Where,

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

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

Therefore, ∆m' = 29 × 1.007825 + 34 × 1.008665 − 62.9296

= 0.591935 u

Mass defect of all the atoms present in the coin, ∆m = 0.591935 × 2.868 × 10^{22}

=1.69766958x10^{22}u.

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

Therefore, ∆m = 1.69766958 × 10^{22} × 931.5 (MeV/c^{2})

Hence, the binding energy of the nuclei of the coin is given as:

E_{b}= ∆mc^{2}

= 1.69766958 × 10^{22} × 931.5(MeV/c^{2})/c^{2}

= 1.581 × 10^{25} MeV

But 1 MeV = 1.6 × 10^{−13} J

E_{b} = 1.581 × 10^{25} × 1.6 × 10^{−13}

= 2.5296 × 10^{12} J

This much energy is required to separate all the neutrons and protons from the given coin.

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