What will be the ratio of the energy released by 4 kg . . .

Question : What will be the ratio of the energy released by 4 kg of hydrogen at sun by fusion process to 23.5 kg of U(235, 92) in the nuclear reactor by fission process? (Assume energy released per fusion is 26 MeV and that per fission is 200 MeV)

Doubt by Parish 

Solution : 

Consider the fusion of Hydrogen to Helium : 

We know, In the sun four Hydrogen atom combine to form one Helium atom and releases 26 MeV of energy

4H (1,1)→He(4,2)

Mass (m) of Hydrogen = 4 kg = 4000 g

No. of moles of Hydrogen

= (Given Mass)/(Molar Mass)

= m/M

= 4000/1
= 4000

No. of hydrogen atom available in 4 kg of Hydrogen
= No. of moles × Avogadro Number

= 4000×6×10²³
=24000×10²³

=24×10²⁶

No. of fusions = (No. of hydrogen atom available in 4 kg of Hydrogen atom)/4
=(24×10²⁶)/4
= 6×10²⁶

Total energy released by 4 kg of Hydrogen in fusion reaction 

=6×10²⁶×26MeV

E₁=156×10²⁶ MeV — (1)

Consider the fission of Uranium 

We know, fission of one atom of U(235,92) produces 200 MeV of energy

Mass (m) of U(235,92) = 23.5 kg = 23500 g

No. of moles of Uranium  

= (Given Mass)/(Molar Mass)

= m/M

= 23500/235
= 100

No. of Uranium atom available in 23500 g of Hydrogen
= No. of moles × Avogadro Number

= 100×6×10²³
=6×10²⁵

No. of fusions = (No. of Uranium atom available in 23500 g of Uranium atom)/1
=6×10^25

Total energy released by 23500 g (23.5 kg) of Uranium in fusion reaction 

=6×10²⁵×200MeV

E₂=120×10²⁶ MeV—(2)

Dividing equation (2) by (1)

E₂/E₁=(120×10²⁶)/(156×10²⁶)
E₂/E₁=120/156

E₂/E₁ = 10/13
E₂:E₁=10:13

E₁:E₂ = 13:10

Similar Question : 

Calculate a compare the energy released by (a) fusion of 1.0 kg of hydrogen deep within the sun and (b) the fission of 1.0 kg of U(235) in a fission reactor.