In the preceding example, at the final speed, 0.9997 c, what was the particle energy as a multiple of the rest energy$m{c}^2$ ? (That is, if it was twice $m{c}^2$, write$m{c}^2$ .) What was the kinetic energy as a multiple of$m{c}^2$? Was the kinetic energy large or small compared to the rest energy? At low speeds, is the kinetic energy large or small compared to the rest energy?

The speed ratio to the speed of light is $\frac{v}{c}=0.9997$.

The energy that an object has as a result of motion is known as kinetic energy. It is described as the effort required to move a mass-determined body from rest to the indicated velocity. The body keeps the kinetic energy it acquired throughout its acceleration unless its speed changes.

The electron system has two kinds of energy, the kinetic energy which associated with its motion and the rest energy where it has zero speed. The summation of both energies called its particle energy and it is given by

$E_{Particle}=\gamma m{c}^2$.............(1)

where m is the mass of the electron, c is the speed of light which equals $3\times {10}^{8}m/s$and $\gamma$is the relativistic factor represents the difference between the energy at the motion and at rest and it is given by

$\gamma =\frac{1}{\sqrt{1-{\left(\frac{v}{c}\right)}^2}}$...............(2)

So, equation (1) will be in the form

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

Now we can plug our values for c, m and $\frac{v}{c}$into equation (2) to get the particle energy of the electron

$$\begin{gathered} E_{Particle}=\frac{m{c}^2}{\sqrt{1-{\left({\displaystyle \frac{v}{c}}\right)}^2}} \\ {E}_{Particle}=\frac{m{c}^2}{\sqrt{1-{\left(0.9997\right)}^2}} \\ {E}_{Particle}=40.8m{c}^{2}J \end{gathered}$$

where the term$m{c}^2$ is the rest energy.

The particle energy represents the total energy at rest plus the energy at motion, so the kinetic energy which is the energy of the electron at motion is given by

$K=E_{Particle}-E_{rest}$..............(3)

Now let us plug our results for $E_{Particle}$and $E_{rest}$into equation (3) to get the kinetic energy,

$$\begin{gathered} K=E_{particle}-E_{rest} \\ K=40.8m{c}^{2}J-m{c}^{2}J \\ K=39.8m{c}^{2}J \end{gathered}$$

As shown, the kinetic energy is larger than the rest energy by 39.8 times. But at low speeds, the kinetic energy is smaller than the rest energy.

$E_{Particle}=40.{8\mathrm{mc}}^2$and $K=39.8m{c}^{2}J$

Therefore, the kinetic energy is larger than the rest energy at high speed while at low speed the kinetic energy is smaller than the rest energy.