91Ó°ÊÓ

The particle’s momentum magnitude is mc.

The Lorentz factor depends only on velocity and not on the particle’s mass and it is expressed as

$\mathit{γ}\mathbf{=}\frac{\mathbf{1}}{\sqrt{\mathbf{1}\mathbf{-}{\mathbf{β}}^{\mathbf{2}}}}$

$\mathbf{β}$is called the speed parameter which is ratio of speed of particle to speed of light.

.$\mathbf{β}\mathbf{}\mathbf{=}\mathbf{}\mathbf{V}\mathbf{/}\mathbf{C}$

It is given that a particle’s momentum magnitude is mc . The relativistic momentum relation is given by

$p=\mathrm{γ}mv$

Putting the given momentum in above expression, we get

$$\begin{gathered} mc=\mathrm{γ}mv \\ \mathrm{γ}v=c \\ \mathrm{γ}\frac{v}{c}=1 \\ \mathrm{γ}\mathrm{β}=1 \end{gathered}$$ … (1)

We know that the Lorentz factor depends only on velocity and not on the particle’s mass and it is expressed as

$\mathrm{γ}=\frac{1}{\sqrt{1-{\mathrm{β}}^2}}$

Inserting this in equation (1), we get

$$\begin{gathered} \frac{\mathrm{β}}{\sqrt{1-{\mathrm{β}}^2}}=1 \\ {\mathrm{β}}^2=1-{\mathrm{β}}^2 \\ \mathrm{β}=\frac{1}{\sqrt{2}} \\ =0.707 \end{gathered}$$

Hence the value of $\mathrm{β}$is, 0.707.

Putting the value of$\mathrm{β}$ in the equation (1),

$$\begin{gathered} \mathrm{γ}=\frac{1}{\mathrm{β}} \\ =\frac{1}{0.707} \\ =1.414 \end{gathered}$$

The value of $\mathrm{γ}$is, 0.414.

The relativistic kinetic energy relation is given by

$$\begin{gathered} K=\left(\mathrm{γ}-1\right)m{c}^2=\left(\mathrm{γ}-1\right)E_o \\ \frac{K}{E_o}=\mathrm{γ}-1 \end{gathered}$$

Substitute all value in the above equation

$$\begin{gathered} \frac{K}{E_o}=1.414- \\ =0.414 \end{gathered}$$

Hence the value of $\frac{K}{E_o}$is,0.414 .

And thus, all the required values are determined.