0
$\begingroup$

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.

$\endgroup$

1 Answer 1

0
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.