How to convert the electron volt into Velocity unit? 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. 
 A: 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 $$K=\frac{mv^2}{2}.$$
A: 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%).
