# Calculating the Energy released in Fusion between Deuterium and Tritium

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.