Assuming that about 200 MeV energy is released per fission of ₉₂U²³⁵ nuclei,
would be the mass of U²³⁵ consumed per day in the fission of reactor of power 1 MW
approximately?
a. 10 kg b. 100 kg c. 1 g d. 10⁻² g
Answers
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1 eV = 1.602×10⁻¹⁹ J
Energy released from fission of 1 nucleus = 200 MeV
200 MeV = 200×10⁶×1.602×10⁻¹⁹ J = 3.204×10⁻¹¹ J
Energy released from 1 mole Uranium = 6.022×10²³×3.204×10⁻¹¹ J
= 1.929×10¹³ J
= 1.929×10⁷ MJ
Power = 1MW = 1 MJ per second
In 1 day, energy produced = 24×3600×1 MJ = 86400 MJ
1mole produces = 1.929×10⁷ MJ
86400 mJ will be produced = 86400/1.929×10⁷ = 0.004479 mole
mass of 1 mole Uranium = 235g
mass of 0.004479 mole Uranium = 235×0.004479 = 1.052 g ≈ 1g
So (c) is the correct option.
Energy released from fission of 1 nucleus = 200 MeV
200 MeV = 200×10⁶×1.602×10⁻¹⁹ J = 3.204×10⁻¹¹ J
Energy released from 1 mole Uranium = 6.022×10²³×3.204×10⁻¹¹ J
= 1.929×10¹³ J
= 1.929×10⁷ MJ
Power = 1MW = 1 MJ per second
In 1 day, energy produced = 24×3600×1 MJ = 86400 MJ
1mole produces = 1.929×10⁷ MJ
86400 mJ will be produced = 86400/1.929×10⁷ = 0.004479 mole
mass of 1 mole Uranium = 235g
mass of 0.004479 mole Uranium = 235×0.004479 = 1.052 g ≈ 1g
So (c) is the correct option.
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