91Ó°ÊÓ

In this problem, we are given that the three particles have same kinetic energy and we are supposed to determine the Lorentz factor and speed parameter for each particle. Some data that is given in the problem is as follows

• The kinetic energy of all the three particles is$K=10.00MeV$ .
• The rest mass energy of the electron is $E_o=0.510998MeV$.
• The rest mass energy of the proton is$E_o=938.272MeV≈938.3MeV$ .
• The rest mass energy of the $\mathrm{Î\pm }$particle is$E_o=3727.40MeV≈3727MeV$

The result of 2^nd postulate of the special theory of relativity is that the clocks run slower for a moving object when measured from a rest frame. The factor by which the clock is running differently is called the Lorentz factor.

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.

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

The relativistic kinetic energy relation is given by

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

Where E₀ is the rest mass energy.

Finding Lorentz factor,

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

Substitute all the value in the above equation.

$$\begin{gathered} \mathrm{γ}=\frac{10.00MeV}{0.510998MeV}+1 \\ =20.57 \end{gathered}$$

Find the speed parameter,

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

Substitute all the value in the above equation.

$$\begin{gathered} \mathrm{β}=\sqrt{1-\frac{1}{{\left(20.57\right)}^2}} \\ =0.9988 \end{gathered}$$

Hence for the electron is:

1. $\mathrm{γ}=20.57$
   
2. $\mathrm{β}=0.9988$
   

$\mathrm{β}=0.1451$

The relativistic kinetic energy relation is given by

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

Where E₀ is the rest mass energy.

Finding Lorentz factor,

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

Substitute all the value in the above equation.

$$\begin{gathered} \mathrm{γ}=\frac{10.00MeV}{938.3MeV}+1 \\ =1.0107 \end{gathered}$$

Find the speed parameter,

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

Substitute all the value in the above equation.

$$\begin{gathered} \mathrm{β}=\sqrt{1-\frac{1}{{\left(1.0107\right)}^2}} \\ =0.1451 \end{gathered}$$

Hence for the proton is:

1. $\mathrm{γ}=1.0107$
   
2. 
   

The relativistic kinetic energy relation is given by

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

Where E₀ is the rest mass energy.

Finding Lorentz factor,

role="math" localid="1663047166810" $$\begin{gathered} K=\left(\mathrm{γ}-1\right)E_o \\ \mathrm{γ}-1=\frac{K}{E_o} \end{gathered}$$

Substitute all the value in the above equation.

$$\begin{gathered} \mathrm{γ}=\frac{10.00MeV}{3727MeV}+1 \\ =1.1478 \end{gathered}$$

Find the speed parameter,

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

Substitute all the value in the above equation.

$$\begin{gathered} \mathrm{β}=\sqrt{1-\frac{1}{{\left(1.1478\right)}^2}} \\ =0.4909 \end{gathered}$$

Hence for the particle is:

1. $\mathrm{γ}=1.1478$
   
2. $\mathrm{β}=0.4909$
   

Thus, the Lorentz factor and the speed parameter is determined for each particle.