I'm trying to calculate the Energy you would get in a fusion reactor from the fusion of deuterium and tritium:
${}^2H+{}^3H \rightarrow {}^4He + n$

Using this Equation:
$E = E_{rest} + E_{kin} = mc^2 + \frac12mv^2$

And these values i found online:
$m_{Deuterium} \approx 2.01410177811u$
$m_{Tritium} \approx 3.01604928u$
$m_{Helium4} \approx 4.002603254u$
$m_{Neutron} \approx 1.03352196257794u$

These velocities are at ~100 million Kelvin
$v_{Deuterium} \approx 1500\frac{km}s$
$v_{Tritium} \approx 1000\frac{km}s$

Plugging in the values i get this:
$E_{Deuterium} \approx 1882.3819988MeV$
$E_{Tritium} \approx 2819.97477352MeV$
$E_{Helium4} \approx 3728.40131MeV$
$E_{Neutron} \approx 962.719610361MeV$

Then the Energy before the reaction minus the energy after the reaction is:
$\Delta E \approx 10.84669MeV$

But on the Wikipedia about fusion it says that the reaction should release $17.59MeV$ in kinetic energy.
I assume the problem could be the inaccurate velocities, but I'm not sure the difference would be so big.


1 Answer 1


Kinetic energies do not have to be taken into account: they only serve to overcome Coulomb repulsion. In the Sun, fusions occur at "low temperature". Do the usual Q checkup with the masses and you'll easily find the correct answer.

  • $\begingroup$ I don't know what you mean by "the usual Q checkup", all I could find was the Q-value, being the energy gain/loss, which I'm calculating here, or the fusion energy gain factor, which I don't know how to solve my problem with. Could you maybe provide a link on what it is and how to calculate it? $\endgroup$
    – BlueSheep3
    Commented May 31, 2022 at 17:09
  • $\begingroup$ Yes , it is the usual Q-value .But use the correct value for neutron mass and you wiill find the correct result by yourself . neutron mass : 1.008 664 915 95 u $\endgroup$ Commented Jun 1, 2022 at 5:53

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