Approximately how much uranium (in kg) must undergo fission per day to provide 1000 MW of power? (Assume an efficiency of 30%).The nuclear reaction is: . m(n) = 1.008665 u
M(U) = 235.043915 u
M(Ba) = 140.9139 u
M(Kr) = 91.8973 u
U = 1.66 * 10 -27 kg
A)1.0
B)3.5
C)2.3
D)4.6
E)0.1

Question

Approximately how much uranium (in kg) must undergo fission per day to provide 1000 MW of power? (Assume an efficiency of 30%).The nuclear reaction is: .  m(n) = 1.008665 u
M(U) = 235.043915 u
M(Ba) = 140.9139 u
M(Kr) = 91.8973 u
U = 1.66 * 10 -27 kg
A)1.0 
B)3.5 
C)2.3 
D)4.6 
E)0.1

Quizplus · Accepted Answer

First, we can use the formula for power to find the amount of energy produced per second:
Power = Energy / Time
1000 MW = Energy / 1 second
Energy = 1 x 10^9 J

Next, we need to find the energy released per fission reaction. From the given nuclear reaction, we can see that one neutron (n) is absorbed by a uranium-235 nucleus (U) to produce two smaller nuclei (Ba and Kr), as well as three neutrons:
U + n → Ba + Kr + 3n

The energy released per fission is approximately 200 MeV (million electron volts), or 3.2 x 10^-11 J (this value can be looked up or calculated from nuclear physics principles).

Now we can find the number of fissions per second required to produce 1 x 10^9 J of energy:
(1 x 10^9 J) / (3.2 x 10^-11 J/fission) = 3.125 x 10^19 fissions/second

Finally, we need to find the mass of uranium-235 required for this many fissions per second. The atomic mass of uranium-235 is 235.043915 u, or 3.90 x 10^-25 kg. We can use the equation:
mass (kg) = (number of fissions/second) x (energy per fission) / (efficiency x mass per fission)
mass (kg) = (3.125 x 10^19 /s) x (3.2 x 10^-11 J/fission) / (0.3 x 235.043915 u x 1.66 x 10^-27 kg/u)
mass (kg) ≈ 3.5 kg/day

Therefore, the answer is B. Approximately 3.5 kg of uranium-235 must undergo fission per day to provide 1000 MW of power with 30% efficiency.

Approximately How Much Uranium (In Kg) Must Undergo Fission Per