# If 200 MeV energy is released per fission of U-235 nucleus then find the mass of U-235 consumed per day in a reactor of power 1MW assuming its eficiency is 80%.

Arun
25763 Points
2 years ago
Assuming that about 200MeV of energy is released per fission of . 92U235 nuceli, the mass of U235 consumed per day in a fission ractor of power 1 megawatt will be approximately . (2356×1023)(27×1020)=1.058g=1g
Sayantan Garai
117 Points
2 years ago
Energy released per disintegration =200 MeV= 200×(1.6×10 ^−13)J=3.2×10^−11 J
Total Energy required =1 MW= 10^6 J/s
No. of integrations required =N= 10^6 / 3.2×10^-11 = 3.125 × 10^16.

Moles (n) = N / Avocado's N = 5.188×10−8 moles/s

Mass consumption =nM= 1.22×10^−5
gms/s
Total consumption in a day =1.22×10 ^−5
gm/s×(24×3600)=1.053 gms

Efficiency = 80%.

Hence, required answer = 1.053×100/80 = 1.31gm.
Yashraj
915 Points
2 years ago
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