Question

How fast does a 2 MeV fission neutron travel through a reactor core?

Answer 1

Answer:

The answer is $1.956 \times 10^7\ m/s$

Explanation:

The amount of energy is not enough to apply the relativistic formula of energy $E = mc^2$, so the definition of energy in this case is

$E = \frac{1}{2}m v^2$.

From the last equation,

$v= \sqrt{2E/m}$

where

$E = 2 MeV = 3.204 \times 10^{-13} J$

and the mass of the neutron is

$m = 1.675\times 10^{-27}\ Kg$.

Then

$v = 1.956 \times 10^7\ m/s$

the equivalent of $0.065$ the speed of light.