Answer to Question #87754 in Mechanics | Relativity for Nicholas Legault

Question

A neutron in a particle accelerator moves at a speed of 0.65c. What is its total energy in the laboratory frame of reference?

Answer 1

$$E=\frac{mc^2}{\sqrt{1-(\frac{v}{c})^2}}$$

$$E=\frac{(1.67 \cdot 10^{-27})(3 \cdot 10^8)^2}{\sqrt{1-0.65^2}}=2.0 \cdot 10^{-10} J$$