How Does E=mc2 Explain Energy Release in Nuclear Fission?

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SUMMARY

The discussion focuses on the application of Einstein's equation E=mc² to nuclear fission, specifically using the reaction of Uranium-235 (92U235) undergoing fission to produce Rubidium-90 (37Rb90) and Cesium-143 (55Cs143). The calculated mass loss during this reaction is 0.18447 AMU, which corresponds to an energy release of 171.74 MeV. In contrast, typical chemical reactions, such as the combustion of methane, release energy on the order of a few electron volts, highlighting the significant difference in energy production between nuclear fission and chemical reactions.

PREREQUISITES
  • Understanding of nuclear fission reactions, specifically involving Uranium-235.
  • Familiarity with atomic mass units (AMU) and their conversion to energy (MeV).
  • Knowledge of basic chemical reactions and energy release, particularly fossil fuel combustion.
  • Ability to perform unit conversions between electron volts and joules.
NEXT STEPS
  • Research the detailed process of nuclear fission and its applications in nuclear reactors.
  • Learn about the conversion of AMU to energy using Einstein's equation E=mc².
  • Explore the differences between nuclear and chemical reactions in terms of energy release.
  • Study the energy content of various fuels, comparing fossil fuels to nuclear energy sources.
USEFUL FOR

Students studying nuclear physics, educators teaching energy concepts, and professionals in the fields of nuclear engineering and chemistry seeking to understand the fundamental differences in energy release mechanisms.

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1. Homework Statement [/b]

The overall question is:
For this assignment explain how the E=mc2 equation applies to nuclear fission. In your answer illustrate your explanation with an example, being sure to distinguish between mass and mass number, and explain how a nuclear equation differs from a chemical equation. In addition compare the energy released during fission with energy produced from a typical chemical reaction (such as fossil fuel oxidation). It may be useful for you to consider that the combustion of methane releases 50.1 kJ/g - how much mass is lost to produce 50.1 kJ?

I put it all here so hopefully someone can make sense of my anwer below. I welcome any suggestions!

3. The Attempt at a Solution [/b]

Started by inventing a reaction: 92U235 + 0n1 -> 37Rb90 + 55Cs143 + 3 0n1. (typical of a nuclear fission reaction.) The actual atomic masses of these are:
Rb90 = 89.91481
Cs143 = 142.92732.
2 0n1 = 2.01732
Sum = 234.85945. Now subtract:
U235 = 235.04392 to get
loss = 0.18447. One AMU = 931 Mev, so 171.74 Mev liberated by the fission. Chemical reactions produce energies on the order of a few electron volts.

This is where I stop...do I need a table of energy conversion values to convert AMU per mole to joules per mole, and then do a bit of arithmetic? Thanks for helping me out!
 
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You already have the energy released (MeV) per U235 atom. The energy release per mole is that times Avogadro's number. Once you have the result in MeV, you don't need a table. 1 electron volt = 1.60217646 × 10^(-19) joules. I just pasted that out of Google.
 

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