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 fissionor by
    • Combining 2 lighter nuclei to form a heavy nucleus known as nuclear fusion.
  • 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 (2963Cu) atoms (of mass 62.92960 u).

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

Atomic mass of (2963Cu) atom, m = 62.92960 u

The total number of (2963Cu) atoms in the coin, N = (NA x m’)/ (Mass number)

Where,

NA = Avogadro’s number = 6.023 × 1023 atoms /g

Mass number = 63 g

Therefore N = (6.023 × 1023 x3)/ (63) =2.868x1022 atoms.

(2963Cu) nucleus has 29 protons and (63-29) = 34 neutrons.

Therefore, Mass defect of this nucleus, ∆m' = 29 × mH + 34 × mn − m

Where,

Mass of a proton, mH = 1.007825 u

Mass of a neutron, mn = 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 × 1022

=1.69766958x1022u.

But 1 u = 931.5 (MeV/c2)

Therefore, ∆m = 1.69766958 × 1022 × 931.5 (MeV/c2)

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

Eb= ∆mc2

= 1.69766958 × 1022 × 931.5(MeV/c2)/c2

= 1.581 × 1025 MeV

But 1 MeV = 1.6 × 10−13 J

Eb = 1.581 × 1025 × 1.6 × 10−13

= 2.5296 × 1012 J

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

 

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