An electron has mass $\mathbf{9}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{31}}\mathbf{kg}$. If the electron's speed is 0.988c.(that is,v/c=0.988), what is its particle energy? What is its rest energy? What is its kinetic energy?

The given data is listed below:

• The mass of the electron is,$m=9\times {10}^{31}\mathrm{kg}$
• The speed of the electron is,$v=0.988c$
• The ratio of the speed of electron and speed of light is,$\frac{v}{c}=0.988$

The kinetic energy of an electron is connected with its motion, while the rest energy is associated with its zero speed. Its particle energy is the total of both energies, and it is given by,

${\mathit{E}}_{\mathbf{p}\mathbf{a}\mathbf{r}\mathbf{t}\mathbf{i}\mathbf{c}\mathbf{l}\mathbf{e}}\mathbf{=}\mathit{\gamma }\mathit{m}{\mathit{c}}^{\mathbf{2}}$ …(i)

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

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

Rearrange equation (i) by substituting $\frac{1}{\sqrt{1-{\left({\displaystyle \frac{v}{c}}\right)}^2}}$for $\gamma$.

$E_{particle}=\frac{mc}{\sqrt{1-{\left(\frac{v}{c}\right)}^2}}$

Substitute all the values in the above equation.

$$\begin{gathered} E_{particle}=\frac{\left(9\times {10}^{-31}kg\right)\left(3\times {10}^{8}m/s\right)}{\sqrt{1-{\left(0.988\right)}^2}} \\ =5.24\times {10}^{-13}\mathrm{kg}.{\mathrm{m}}^2/{\mathrm{s}}^2\times \frac{1\mathrm{J}}{1\mathrm{kg}.{\mathrm{m}}^2/{\mathrm{s}}^2} \\ =5.24\times {10}^{-13}\mathrm{J} \end{gathered}$$

Thus, the particle energy of the electron is $5.24\times {10}^{-13}\mathrm{J}$.

It is known that the rest energy is the energy when the speed of the electron is V=0 , the factor is$\gamma \approx 1$ . So, the expression for the rest energy is as follows,

$E_{rest}={\mathrm{mc}}^2$

Substitute all the values in the above expression.

$$\begin{gathered} E_{rest}=\left(9\times {10}^{-31}\mathrm{kg}\right){\left(3\times {10}^8\mathrm{m}/\mathrm{s}\right)}^2 \\ =8.1\times {10}^{-14}\mathrm{kg}.{\mathrm{m}}^2/{\mathrm{s}}^2\times \frac{1\mathrm{J}}{1\mathrm{kg}.{\mathrm{m}}^2/{\mathrm{s}}^2} \\ =8.1\times {10}^{-14}\mathrm{J} \end{gathered}$$

Thus, the rest energy of the electron is$8.1\times {10}^{-14}\mathrm{J}$ .

The expression for the kinetic energy is expressed as follows,

$K=E_{particle}-E_{rest}$

Substitute all the values in the above expression.

$$\begin{gathered} K=5.24\times {10}^{-13}\mathrm{J}-8.1\times {10}^{-14}\mathrm{J} \\ =4.43\times {10}^{-13}\mathrm{J} \end{gathered}$$

Therefore, the kinetic energy of the electron is$4.43\times {10}^{-13}\mathrm{J}$ .