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, stafusa Mar 5 at 22:33

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  • $\begingroup$ Don't you know anything about the velocity of the other particles? It seems that you should also use the conservation of momentum. $\endgroup$ – TheAverageHijano Mar 4 at 8:43
  • $\begingroup$ 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. $\endgroup$ – Jon Custer Mar 4 at 19:49
  • 1
    $\begingroup$ Which particle? $\endgroup$ – Bill N Mar 5 at 22:03

You can’t convert it to a velocity, because it isn’t the kinetic energy of one thing. It is the total kinetic energy of four different fission products, each moving independently with its own velocity. To find the velocity of each one, you would need to know how this total kinetic energy is divided between the four fission products.