How to Calculate Energy Released in Nuclear Fusion

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SUMMARY

The discussion focuses on calculating the energy released during the nuclear fusion of tritium and deuterium, resulting in helium and a neutron. The formula used is (Σ mass (reactants) - Σ mass (products)) x 931.5 MeV/amu, which yields a theoretical energy release of approximately 17.58 MeV. The user initially calculated 18.1 MeV due to using atomic masses instead of nuclear masses for tritium. The correction highlights the importance of using precise nuclear mass values in fusion energy calculations.

PREREQUISITES
  • Understanding of nuclear fusion principles
  • Familiarity with mass-energy equivalence and binding energy calculations
  • Knowledge of atomic and nuclear mass values
  • Basic proficiency in using MeV (mega-electronvolts) as a unit of energy
NEXT STEPS
  • Research the mass of nuclei for various isotopes, including tritium and deuterium
  • Learn about the role of binding energy in nuclear reactions
  • Explore the implications of mass-energy equivalence in nuclear physics
  • Study the differences between atomic mass and nuclear mass
USEFUL FOR

Students and researchers in nuclear physics, educators teaching nuclear fusion concepts, and anyone interested in the calculations related to energy release in fusion reactions.

frazzle96
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Hi,

I'm doing some coursework on nuclear fusion, and am trying to calculate the energy released by fusing a tritium and a deuterium atom together. I used the equation:

( Σ mass (reactants) - Σ mass (products) ) x 931.5 MeV/amu = Binding energy.

and Deuterium + Tritium = Helium + neutron

I know the value should be about 17.58 MeV, however when I did it it came out as about 18.1.

The values I used are Triton: 3.0160492u , Deuteron: 2.0135532u , Helium nucleus: 4.0015061u and a neutron: 1.0086649u.

Thanks in advance

Frazer
 
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You're off by about 0.5 MeV, which is the mass of an electron. Are the values you quoted the mass of the atoms or the nuclei?
 
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ah, I think the value I used for tritium was the atomic mass rather than the mass of the nuclei... thanks very much!
 

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