Question: What is$\mathit{\beta }$ for a particle with (a) $\mathit{K}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{00}{\mathit{E}}_{\mathbf{o}}$and (b)$\mathit{E}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{00}{\mathit{E}}_{\mathbf{o}}$ ?

We are asked to determine the speed parameter for two cases, where in one kinetic energy is given and in the other case total energy is given.

1. The kinetic energy of the particle is$K=2.00E_o$ .
2. The total energy of the particle is $E=2.00E_o$.

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

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

is called the speed parameter which is ratio of speed of particle to speed of light.

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

The relativistic kinetic energy relation is given by

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

Substituting the given value of in the above equation

$$\begin{gathered} 2.00E_o=\left(\gamma -1\right)E_o \\ \gamma -1=2.00 \\ \gamma =3.00 \end{gathered}$$

Finding the speed parameter,

$$\begin{gathered} \frac{1}{\sqrt{1-{\beta }^2}}=3 \\ \beta =\sqrt{1-\frac{1}{3^2}} \\ =\sqrt{0.89} \end{gathered}$$

Hence the $\beta is0.943$ .

The relativistic Total energy relation is given by

$E=\gamma m{c}^2=\gamma E_o$

Substituting the given value of E in the above equation

$$\begin{gathered} 2.00E_o=\gamma E_o \\ \gamma =2.00 \end{gathered}$$

Finding the speed parameter,

$$\begin{gathered} \frac{1}{\sqrt{1-{\beta }^2}}=2 \\ \beta =\sqrt{1-\frac{1}{2^2}} \\ =\sqrt{0.75} \\ =0.866 \end{gathered}$$

Hence the$\beta is0.866$

The speed parameter for each case is thus found.