91Ó°ÊÓ

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 $E=2.00E_o$.

.

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

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

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

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

The relativistic kinetic energy relation is given by

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

Substituting the given value of in the above equation

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

Finding the speed parameter,

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

Hence the $\mathrm{β}is0.943$ .

The relativistic Total energy relation is given by

$E=\mathrm{γ}m{c}^2=\mathrm{γ}{E}_o$

Substituting the given value of E in the above equation

$$\begin{gathered} 2.00E_o=\mathrm{γ}{E}_o \\ \mathrm{γ}=2.00 \end{gathered}$$

Finding the speed parameter,

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

Hence the$\mathrm{β}is0.866$

The speed parameter for each case is thus found.