Question

Calculate the kinetic energy required to accelerate a neutron from a rest position to 0.99990 If the mass of the neutron is 1.68*10^ -27 kg

Answer 1

The relativistic kinetic energy is given as follows:

$$K = mc^2\left( \dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}} - 1 \right)$$

where $m = 1.68\times10^{-27}kg$ is the mass of the neutron, $c = 3\times 10^8 m/s$ is the speed of light, $v = 0.99990c$ is the speed of the neutron. Thus, obtain:

$$K = 1.68\times10^{-27}\cdot (3\times 10^8)^2\left( \dfrac{1}{\sqrt{1-\dfrac{(0.99990c)^2}{c^2}}} - 1 \right) \approx 1.05\times 10^{-8}J$$

Answer. $1.05\times 10^{-8}J$.