Question: A certain particle of mass m has momentum of magnitude mc .What are (a) $\mathit{\beta }$, (b)$\mathit{\gamma }$, and (c) the ratio$\mathit{K}\mathbf{/}{\mathit{E}}_{\mathbf{0}}$?

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

$\mathit{\gamma }\mathbf{=}\frac{\mathbf{1}}{\sqrt{\mathbf{1}\mathbf{-}{\mathbf{\beta }}^{\mathbf{2}}}}$

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

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

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

$p=\gamma mv$

Putting the given momentum in above expression, we get

$$\begin{gathered} mc=\gamma mv \\ \gamma v=c \\ \gamma \frac{v}{c}=1 \\ \gamma \beta =1 \end{gathered}$$ … (1)

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

$\gamma =\frac{1}{\sqrt{1-{\beta }^2}}$

Inserting this in equation (1), we get

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

Hence the value of $\beta$is, 0.707.

Putting the value of$\beta$ in the equation (1),

$$\begin{gathered} \gamma =\frac{1}{\beta } \\ =\frac{1}{0.707} \\ =1.414 \end{gathered}$$

The value of $\gamma$is, 0.414.

The relativistic kinetic energy relation is given by

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

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.