Calculating Energy of Fission in Deuterium-Tritium Reaction

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The discussion focuses on calculating the energy released during the fission of deuterium and tritium nuclei to form helium. The relevant nuclear reaction is presented, and the mass-energy equivalence formula, ΔE=Δm c², is applied. The calculated mass defect is 0.018334 u, leading to an energy release of approximately 17.08 MeV. However, discrepancies arise when comparing this result to textbook values, which suggest an energy release of around 17.6 MeV or 17.8 MeV. The discussion highlights the importance of using accurate mass values for precise calculations in nuclear reactions.
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Homework Statement



Find the energy which comes out with fission of the nucleus of deuterium and tritium, so we receive like product the nucleus of helium.

mass of tritiumm(\stackrel{3}{1}H)=3,016049u
mass of the nucleus of deuterium m(\stackrel{2}{1}H)=2,013553u
mass of the nucleus of helium m(\stackrel{4}{2}He)=4,002603u


Homework Equations



\Delta E=\Delta m c^2

The Attempt at a Solution



The nuclear reaction

\stackrel{2}{1}H + \stackrel{3}{1}H \rightarrow \stackrel{4}{2}He + \stackrel{1}{0}n + energy

How will I find the mass of triton( the nucleus of tritium) ?
 
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Physicsissuef said:
How will I find the mass of triton( the nucleus of tritium) ?
Take the mass of the atom and subtract the mass of the electron. The binding energy of the electron is not enough to make a significant difference in the mass of the atom.

AM
 
(3,016049u+2,013553u)-(4,002603u+1,008665u)=0,018334

0,018334*931,494 MeV=17,078011 MeV

And in my textbook results it is 17,8 MeV. Is their fault?
 
The answer should be 17.6MeV, this is well known reaction to me.

And you will obtain that answer if you use m(\stackrel{2}{1}H)=2,014101783u

Edit, link:
http://www.nndc.bnl.gov/amdc/masstables/Ame2003/mass.mas03
 
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