Approximately How Much Uranium (In Kg)must Undergo Fission Per Day

Approximately how much uranium (in kg) must undergo fission per day to provide 1 000 MW of power? (Assume an efficiency of 30%) .The nuclear reaction is $\mathrm { n } + { } _ { 92 } ^ { 235 } \mathrm { U } \rightarrow { } _ { 56 } ^ { 141 } \mathrm { Ba } + { } _ { 36 } ^ { 92 } \mathrm { Kr } + 3 \mathrm { n }$ . m(n) = 1.008 665 u
M(U) = 235.043 915 u
M(Ba) = 140.913 9 u
M(Kr) = 91.897 3 u
U = 1.66 *10 ^{- 27} kg

A) 1.0
B) 3.5
C) 2.3
D) 4.6
E) 0.1

Correct Answer:

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