91Ó°ÊÓ

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

The total energy E of a particle is

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

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

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

$\mathrm{γ}$is the Lorentz factor. The expression of$\mathrm{γ}$ is:

$\mathrm{γ}=\frac{1}{\sqrt{1−v^2/c^2}}$

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

For${\mathrm{Ï\dots }}_1=0.100c$

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

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

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

For${\mathrm{Ï\dots }}_2=0.500c$

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

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

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

For${\mathrm{Ï\dots }}_3=0.900c$

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

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

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

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

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

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

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

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

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

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

Hence, $2.10×{10}^{-11}J$work-done must be required to increase the protons speed from 0.100c to 0.500c, $1.72×{10}^{-10}J$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.