About this question,I found this from the Wikipedia:nuclear fission produces neutrons with a mean energy of 2 MeV (200 TJ/kg, i.e. 20,000 km/s), which qualifies as "fast". For my question, I only knew a few detail that converting Electron volt to Velocity unit,and electron volt is closely related with energy density, but I don' know the conversion formula of electron volt to velocity unit.

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    $\begingroup$ An eV can be converted to other energy units, such as the Joule. If you know the mass of the neutron, you can easily convert to a non-relativistic velocity. $\endgroup$ – Jon Custer Dec 23 '17 at 8:34

Electron volt is an energy unit,1 eV = $1.602\times 10^{-19}$ J. It can be used in the context of the kinetic energy of a particle, the potential energy of a system, the mass-energy of a system or particle and so on. In your situation, it sounds like it is used as a kinetic energy.

From there, you need to determine whether the particle is likely to be moving relativistically or not. If the kinetic energy, $K$, is greater than 10% of the mass energy ($mc^2$), then you should probably use $$K=mc^2\left(\frac{1}{\sqrt{1-v^2/c^2}}-1\right).$$

If $K$ is less than 10%, you can get away with $$K=\frac{mv^2}{2}.$$


For small particles, the energy is sometimes conveniently expressed in eV - the energy an electron gets when accelerated through a 1 Volt field - and that's equivalent to about $1.6\cdot 10^{-19}~\rm{J}$

For the energy level you are talking about, you're in the relativistic regime. At low velocities, you can say the kinetic energy is given by $E=\frac12 mv^2$ from which it would follow that $v=\sqrt{\frac{2E}{m}}$; if that gives you a velocity that is close to (or greater than) the speed of light, you need to make a relativistic adjustment.

The energy is $1.6\cdot 10^{-19}\times 2\cdot 10^6~\rm{J}$, and the mass of the neutron is roughly $1.6\cdot 10^{-27}~\rm{kg}$.

At 2 MeV, your neutron is still OK - the above equation gives 20,000 km/s which is considerably less than the speed of light, 300,000 km/s. So there's no need to make the relativistic correction (which would change the result by about 0.2%).

  • $\begingroup$ Thank you!And as your answer ,when I'm calculating one particle's energy level ,I need to consider one particle 's relativistic effect and mass when it's traveling near the light speed, because the relativistic mass is closely related with its kinetic energy near light speed ,is it correct? $\endgroup$ – Ekpyrotic Branes Dec 24 '17 at 0:42
  • $\begingroup$ Yes that sounds correct. $\endgroup$ – Floris Dec 24 '17 at 1:07

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