Question:

A certain nuclear reactor is designed to produce energy through the process of nuclear fission. In one particular reaction, a uranium-235 nucleus undergoes fission and produces two daughter nuclei, along with several neutrons and a large amount of energy.

a) Calculate the energy released in MeV when a uranium-235 nucleus undergoes fission. (Given: Mass of uranium-235 nucleus = 3.90 x 10^-25 kg, Speed of light, c = 3.00 x 10^8 m/s, 1 MeV = 1.60 x 10^-13 J)

b) If this reaction produces an average of 2.5 neutrons per fission event, calculate the total energy released in a reactor that undergoes 1.0 x 10^21 fission events per second.

Answer:

a) The energy released in MeV when a uranium-235 nucleus undergoes fission can be calculated using Einstein's mass-energy equivalence principle, E=mc^2, where E is the energy released, m is the mass defect, and c is the speed of light. The mass defect (Δm) is the difference between the mass of the uranium-235 nucleus and the combined mass of the daughter nuclei and neutrons produced.

First, we calculate the mass defect:

The mass of the uranium-235 nucleus is given as m = 3.90 x 10^-25 kg.

The mass of the daughter nuclei and neutrons can be assumed to be approximately 4/9 of the original uranium-235 nucleus.

Δm = (4/9)m - m Δm = (-5/9)m

Now, we calculate the energy released using the mass-energy equivalence principle:

E = Δmc^2 E = (-5/9)m(c^2) E = (-5/9)(3.90 x 10^-25 kg)(3.00 x 10^8 m/s)^2 E ≈ 2.48 x 10^-11 J

To convert the energy from joules to MeV:

E(MeV) = E / (1.60 x 10^-13 J/MeV) E(MeV) ≈ (2.48 x 10^-11 J) / (1.60 x 10^-13 J/MeV) E(MeV) ≈ 155 MeV

Therefore, the energy released when a uranium-235 nucleus undergoes fission is approximately 155 MeV.

b) To calculate the total energy released in a reactor that undergoes 1.0 x 10^21 fission events per second, we can use the energy released per fission event and the number of fission events per second.

The energy released per fission event = 155 MeV.

The total energy released per second: Total energy = Energy per fission event * Number of fission events per second Total energy = (155 x 10^6 eV) * (1.0 x 10^21 fission events/s) Total energy ≈ 1.55 x 10^29 eV/s

Converting the total energy to joules: Total energy ≈ (1.55 x 10^29 eV) * (1.60 x 10^-19 J/eV) Total energy ≈ 2.48 x 10^10 J/s

Therefore, the total energy released in the reactor that undergoes 1.0 x 10^21 fission events per second is approximately 2.48 x 10^10 J/s.

Markdown:

a) The energy released when a uranium-235 nucleus undergoes fission is approximately 155 MeV.

b) The total energy released in the reactor that undergoes 1.0 x 10^21 fission events per second is approximately 2.48 x 10^10 J/s.