# How to convert MeV to velocity? [closed]

For example, in the fission process:

$${}^{235}\mathrm{U} + n \rightarrow {}^{140}\mathrm{Xe} + {}^{94}\mathrm{Sr} + 2n$$

the masses in atomic mass units are:

\begin{align} M\left({}^{235}\mathrm{U}\right) &= 235.04393 \\ M(n) &= 1.008665 \\ M\left({}^{94}\mathrm{Sr}\right) &= 93.915361 \\ M\left({}^{140}\mathrm{Xe}\right) &= 139.92164 \end{align}

Therefore, the energy released is:

\begin{align} &\left[M\left({}^{235}\mathrm{U}\right) + M(n) - M\left({}^{94}\mathrm{Sr}\right) - M\left({}^{140}\mathrm{Xe}\right) - M(2n)\right] \times 931.5 \,\mathrm{MeV} \\ &= 0.22251 \times 931.5 \,\mathrm{MeV} \\ &= 208.2 \,\mathrm{MeV} \\ \end{align}

How do I change 208.2 MeV to velocity (m/s)?

## closed as unclear what you're asking by Kyle Kanos, Jon Custer, ZeroTheHero, Bill N, stafusaMar 5 at 22:33

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• Don't you know anything about the velocity of the other particles? It seems that you should also use the conservation of momentum. – TheAverageHijano Mar 4 at 8:43
• Since that energy is going into 4 (or more given late neutrons) particles with widely varying masses going in different directions, it is hard to come up with a single number. – Jon Custer Mar 4 at 19:49
• Which particle? – Bill N Mar 5 at 22:03