What is the kinetic energy of a proton moving at $\mathbf{\left(}\mathbf{a}\mathbf{\right)}\mathbf{}\mathbf{}\mathbf{0}\mathbf{.}\mathbf{100}\mathbf{c}\mathbf{;}\mathbf{}\mathbf{\left(}\mathbf{b}\mathbf{\right)}\mathbf{}\mathbf{}\mathbf{0}\mathbf{.}\mathbf{500}\mathbf{c}\mathbf{;}\mathbf{}\mathbf{\left(}\mathbf{c}\mathbf{\right)}\mathbf{}\mathbf{}\mathbf{0}\mathbf{.}\mathbf{900}\mathbf{c}$? How much work must be done to (d) increase the proton’s speed from 0.100 c to 0.500c and (e) increase the proton’s speed from 0.500 c to 0.900 c ? (f) How do the last two results compare to results obtained in the nonrelativistic limit?

Kinetic energy is the difference between total energy and rest energy.

The total energy E of a particle is

$\begin{array}{r}E=K+m{c}^2\\ =\frac{m{c}^2}{\sqrt{1-v^2/c^2}}\\ =\gamma m{c}^2\end{array}$

Where, K is the relativistic kinetic energy. The expression of K is:

$K=\frac{m{c}^2}{\sqrt{1-v^2/c^2}}-m{c}^2$

$\gamma$is the Lorentz factor. The expression of$\gamma$ is:

$\gamma =\frac{1}{\sqrt{1-v^2/c^2}}$

The energy${\mathbf{mc}}^{\mathbf{2}}$ associated with rest mass rather than motion is known as rest energy of the particle.

For${\upsilon }_1=0.100c$

$\begin{array}{r}{\gamma }_1=\frac{1}{\sqrt{1-v^2/c^2}}\\ =\frac{1}{\sqrt{1-[0.100c{]}^2/c^2}}\\ =1.01\end{array}$

The mass of proton$m=1.67\times {10}^{-27}kg$

$$\begin{gathered} K_1=\left({\gamma }_1-1\right)m{c}^2 \\ =(1.01-1)\left(1.67\times {10}^{-27}\right)\left(3\times {10}^8\right) \\ =1.50\times {10}^{-12}\mathrm{J} \end{gathered}$$

For${\upsilon }_2=0.500c$

$$\begin{gathered} {\gamma }_2=\frac{1}{\sqrt{1-v^2/c^2}} \\ =\frac{1}{\sqrt{1-[0.500c{]}^2/c^2}} \\ =1.15 \end{gathered}$$

The mass of proton$m=1.67\times {10}^{-27}kg$

$$\begin{gathered} K_2=\left({\gamma }_2-1\right)m{c}^2 \\ =(1.15-1)\left(1.67\times {10}^{-27}\right)\left(3\times {10}^8\right) \\ =2.25\times {10}^{-11}\mathrm{J} \end{gathered}$$

For${\upsilon }_3=0.900c$

$$\begin{gathered} {\gamma }_3=\frac{1}{\sqrt{1-v^2/c^2}} \\ =\frac{1}{\sqrt{1-[0.900c{]}^2/c^2}} \\ =2.29 \end{gathered}$$

The mass of proton$m=1.67\times {10}^{-27}kg$

$$\begin{gathered} K_3=\left({\gamma }_3-1\right)m{c}^2 \\ =(2.9-1)\left(1.67\times {10}^{-27}\right)\left(3\times {10}^8\right) \\ =1.94\times {10}^{-10}\mathrm{J} \end{gathered}$$

Hence, the kinetic energy of proton moving at 0.100 c is$1.50\times {10}^{-12}J$ and the kinetic energy of proton moving at 0.500 c is$2.25\times {10}^{-11}J$ and the kinetic energy of proton moving at 0.900c is $1.94\times {10}^{-10}J$.

The work-done to increase to protons from 0.100 c to 0.500 c

$$\begin{gathered} W_1=K_2-K_1 \\ =\left(2.25\times {10}^{-11}\right)-\left(1.50\times {10}^{-12}\right) \\ =2.10\times {10}^{-11}\mathrm{J} \end{gathered}$$

The work-done to increase to protons from 0.500c to 0.900 c

$$\begin{gathered} W_2=K_3-K_2 \\ =\left(1.94\times {10}^{-10}\right)-\left(2.25\times {10}^{-11}\right) \\ =1.72\times {10}^{-10}\mathrm{J} \end{gathered}$$

For nonrelativistic limit the work-done increase with protons speed from 0.100c to 0.500c and 0.500c to 0.900c .

$$\begin{gathered} W_1=\frac{1}{2}\left(1.67\times {10}^{-27}\right)\left[(0.500c{)}^2-(0.100c{)}^2\right]\left(3\times {10}^8\right) \\ =1.80\times {10}^{-11}\mathrm{J} \\ {\mathrm{W}}_2=\frac{1}{2}\left(1.67\times {10}^{-27}\right)\left[(0.900\mathrm{c}{)}^2-(0.500\mathrm{c}{)}^2\right]\left(3\times {10}^8\right) \\ =4.21\times {10}^{-11}\mathrm{J} \end{gathered}$$

Hence, $2.10\times {10}^{-11}J$work-done must be required to increase the protons speed from 0.100c to 0.500c, $1.72\times {10}^{-10}J$work-done must be required to increase the protons speed from 0.500c to 0.900c and the deviation of nonrelativistic results from the relativistic ones will increase with increase in protons speed.