91Ó°ÊÓ

The mass of the electron,$m=9.10938188×{10}^{-31}\text{kg}$

The expression to calculate the Lorentz factor is given as follows.

$\mathit{γ}\mathbf{=}\frac{\mathbf{K}}{\mathbf{m}{\mathbf{c}}^{\mathbf{2}}}\mathbf{+}\mathbf{1}$ …(¾±)

Here, is the kinetic energy of electron, is the speed of the light.

The expression to calculate the speed parameter is given as follows.

$\mathit{β}\mathbf{=}\sqrt{\mathbf{1}\mathbf{-}\frac{\mathbf{1}}{{\mathbf{γ}}^{\mathbf{2}}}}$ …(¾±¾±)

Conversion from kiloelectron volt to joule,

$\mathbf{1}\mathbf{}\mathbf{\text{eV}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{6}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{19}}\mathbf{}\mathbf{\text{J}}$

(a)

The kinetic energy of the electron,

$$\begin{gathered} K=1\text{keV} \\ K=1×{10}^3×1.6×{10}^{-19} \text{J} \\ K=1.6×{10}^{-16}\text{J} \end{gathered}$$

Calculate the Lorentz factor.

Substitute $1.6×{10}^{-16}\text{JforK,}9.10938188×{10}^{-31}\text{kgformand}3×{10}^{8}m/s\text{forc}$into equation (i).

$$\begin{gathered} \mathrm{γ}=\frac{1.6×{10}^{-16}\text{J}}{9.10938188×{10}^{-31}\text{kg}×{\left(3×{10}^{8}m/s\right)}^2}+1 \\ \mathrm{γ}=0.19516×{10}^{-2}+1 \\ \mathrm{γ}=1.0019516 \end{gathered}$$

Hence the value of the Lorentz factor is $1.0019516$.

(b)

Calculate the value of the speed parameter.

Substitute for into equation (ii).

$$\begin{gathered} \mathrm{β}=\sqrt{1-\frac{1}{{1.0019516}^2}} \\ \mathrm{β}=\sqrt{1-0.99610} \\ \mathrm{β}=0.06238432 \end{gathered}$$

Hence the value of speed parameter is $0.06238432$.

(c)

The kinetic energy of the electron,

$$\begin{gathered} K=1\text{MeV} \\ K=1×{10}^6×1.6×{10}^{-19}\text{J} \\ K=1.6×{10}^{-13}\text{J} \end{gathered}$$

Calculate the Lorentz factor.

Substitute$1.6×{10}^{-13}\text{JforK,}9.10938188×{10}^{-31}\text{kgformand}3×{10}^8\text{forc}$ into equation (i).

$$\begin{gathered} \frac{1.6×{10}^{-13}\text{J}}{9.10938188×{10}^{-31}\text{kg}×{\left(3×{10}^{8}m/s\right)}^2}+1 \\ \mathrm{γ}=0.19516955×10+1 \\ \mathrm{γ}=2.9516955 \end{gathered}$$

Hence the value of the Lorentz factor is $2.9516955$.

(d)

Calculate the value of the speed parameter.

Substitute $2.9516955$for $\mathrm{γ}$into equation (ii).

$$\begin{gathered} \mathrm{β}=\sqrt{1-\frac{1}{{2.9516955}^2}} \\ \mathrm{β}=\sqrt{1-0.11477753} \\ \mathrm{β}=0.94086262 \end{gathered}$$

Hence the value of speed parameter is 0.94086262.

(e)

The kinetic energy of the electron,

$$\begin{gathered} K=1\text{GeV} \\ K=1×{10}^9×1.6×{10}^{-19}\text{J} \\ K=1.6×{10}^{-10}\text{J} \end{gathered}$$

Calculate the Lorentz factor.

Substitute$1.6×{10}^{-10}\text{Jfork,}9.10938188×{10}^{-31}\text{kgformand}3×{10}^8\text{forc}$ into equation (i).

$$\begin{gathered} \mathrm{γ}=\frac{1.6×{10}^{-10}\text{J}}{9.10938188×{10}^{-31}\text{kg}×{\left(3×{10}^{8}m/s\right)}^2}+1 \\ \mathrm{γ}=0.19516955×{10}^4+1 \\ \mathrm{γ}=1951.6955 \end{gathered}$$

Hence the value of the Lorentz factor is $1951.6955$.

(f)

Calculate the value of the speed parameter.

Substitute for into equation (ii).

$$\begin{gathered} \mathrm{β}=\sqrt{1-\frac{1}{{1951.6955}^2}} \\ \mathrm{β}=\sqrt{1-2.62×{10}^{-7}} \\ \mathrm{β}=0.99999987 \end{gathered}$$

Hence the value of speed parameter is $0.99999987$.