Calculate mass loss in fission of Uranium-235

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SUMMARY

The fission of Uranium-235 (^235_{92}U) releases approximately 200 MeV of energy, which corresponds to a mass loss of about 0.1% of the original mass. Using Einstein's equation E=mc^2, the mass converted into energy is calculated to be 3.56 x 10^-28 kg. The total mass of the U-235 nucleus is approximately 3.92 x 10^-25 kg, leading to the conclusion that the mass loss during fission is a small but significant fraction. This calculation is essential for understanding the energy yield from nuclear fission reactions.

PREREQUISITES
  • Understanding of Einstein's equation E=mc^2
  • Basic knowledge of nuclear fission processes
  • Familiarity with energy units such as MeV and Joules
  • Ability to perform unit conversions and percentage calculations
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  • Study the principles of nuclear fission and its applications in energy generation
  • Learn about the mass-energy equivalence and its implications in physics
  • Explore the properties and behavior of Uranium-235 in nuclear reactions
  • Investigate the methods for calculating energy release in different nuclear reactions
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xatu
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Problem:

The fission of ^{235}_{92}U releases approximately 200 MeV. What percentages of the orginal mass of ^{235}_{92}U + n disappears?

Solution:

Using E=mc^2 we can estimate the amount of mass m converted into energy E. Solving for m we get,

m=E/c^2=[(2 x 10^8 eV)(1.60 x 10^-19 J/eV)]/[3 x 10^8 m/s]^2=3.56 x 10-28 kg

Certainly a small amount. However, the answer in my text is 0.1%. How can this mass be converted into a percentage when the initial quantity was never given?
 
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xatu said:
However, the answer in my text is 0.1%. How can this mass be converted into a percentage when the initial quantity was never given?
You'll need to calculate the approximate mass of the U-235 nucleus.
 
That makes sense. I got the right answer, too.

Total mass is ≈ (1.67 x 10^-27 kg)*235 = 3.92 x 10^-25 kg. So the fraction of mass converted into energy is (3.56 x 10^-28 kg)/(3.92 x 10^-25)*100% ≈ 0.1%.

Thanks.
 

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